MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 6 - MCQExams.com
CBSE
Class 12 Medical Physics
Dual Nature Of Radiation And Matter
Quiz 6
Calculate the ratio of the energy of the photon $$A$$ to the energy of the photon $$B$$ if frequency of the photon $$A$$ is twice the frequency of the photon $$B$$.
Report Question
0%
4
0%
2
0%
1
0%
$$\frac{1}{2}$$
0%
$$\frac{1}{4}$$
Explanation
Since the energy of a photon is proportional to its frequency, twice the frequency will give twice the energy.
Find out the frequency of a photon which has $$6.6 \times {10}^{-18} J$$
of energy. Planck's constant, $$h$$
, is $$6.6 \times {10}^{-34} J s$$.
Report Question
0%
$$1.0 \times {10}^{-52} Hz$$
0%
$$1.0 \times {10}^{-16} Hz$$
0%
$$1.0 Hz$$
0%
$$1.0 \times {10}^{16} Hz$$
0%
$$1.0 \times {10}^{52} Hz$$
Explanation
Given : $$h = 6.6\times 10^{-34}$$ $$Js$$
Energy of photon $$E = h\nu$$
$$\therefore$$ $$6.6\times 10^{-18} = 6.6\times 10^{-34} \times \nu$$ $$\implies \nu = 1.0 \times 10^{16}$$ $$Hz$$
A photon can eject an electron from the surface of a photo-voltaic metal. Identify which of the following is must condition for above statement.
Report Question
0%
The frequency of the photon is above the activation minimum
0%
The wavelength of the photon is above the activation minimum
0%
The speed of the photon is above the activation minimum
0%
The momentum of the photon is below the activation minimum
0%
The momentum of the impacted electron is above the activation minimum
Explanation
According to Einstein's Photoelectric effect, the maximum kinetic energy of an ejected electron when a photon of energy $$h\nu$$ strikes a metal of work potential $$W_0$$ is $$KE=h\nu-W_0$$
Thus for photoelectric effect to take place,
$$KE>0$$
$$\implies h\nu-W_0>0$$
$$\implies \nu>\dfrac{W_0}{h}$$
By means of the diffraction experiment, it is determined that the electron's de Broglie wavelength is $$6.6 \times {10}^{-10} m$$. What is the electron's linear momentum? Use Planck's constant, $$h = 6.6 \times {10}^{-34} J{A} s$$.
Report Question
0%
$$1.0 \times {10}^{-44} kg\, {A} {m}/{s}$$
0%
$$1.0 \times {10}^{-24} kg\, {A} {m}/{s}$$
0%
$$1.0 \times {10}^{24} kg \,{A} {m}/{s}$$
0%
$$2.0 \times {10}^{24} kg\, {A} {m}/{s}$$
0%
$$1.0 \times {10}^{44} kg \,{A} {m}/{s}$$
Explanation
From de Broglie hypothesis , the linear momentum, $$p=\dfrac{h}{\lambda}=\dfrac{6.6\times 10^{-34}}{6.6\times 10^{-10}}=10^{-24} kg A^o m/s$$
__________ is the wavelength of photon of energy $$35$$KeV.
$$h=6.625\times 10^{-34}$$J-s, $$c=3\times 10^8$$m/s, $$1$$eV$$=1.6\times 10^{-19}$$J.
Report Question
0%
$$35\times 10^{-12}$$m
0%
$$35\overset{o}{A}$$
0%
$$3.5$$nm
0%
$$3.5\overset{o}{A}$$
Explanation
Energy of photon $$E = 35 \ KeV$$
$$\therefore$$ $$E = 35\times 1000\times 1.6\times 10^{-19} \ J = 5.6\times 10^{-15} \ J$$
Wavelength of photon $$\lambda = \dfrac{hc}{E}$$
$$\implies \ \lambda = \dfrac{6.625\times 10^{-34}\times 3\times 10^8}{5.6\times 10^{-15}} = 35\times 10^{-12} \ m$$
A particle of mass M at rest decays into two masses $$\displaystyle { m }_{ 1 }$$ and $$\displaystyle { m }_{ 2 }$$ with non zero velocities. The ratio of de-Broglie wavelengths of the particles $$\displaystyle \frac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } $$ is:
Report Question
0%
$$\displaystyle \frac { { m }_{ 2 } }{ { m }_{ 1 } } $$
0%
$$\displaystyle \frac { { m }_{ 1 } }{ { m }_{ 2 } } $$
0%
$$\displaystyle \frac { \sqrt { { m }_{ 1 } } }{ \sqrt { { m }_{ 2 } } } $$
0%
1:1
Explanation
Initial momentum of the particle is zero as it is at rest.
Thus according to law of conservation of momentum, the final momentum of the system must be zero.
Let the momentum of the two fragments be $$p_1$$ and $$p_2$$.
$$\implies$$ $$p_1 = p_2$$
de-Broglie wavelength $$\lambda = \dfrac{h}{p}$$
where $$p$$ is the momentum of the particle and $$h$$ is Planck's constant.
$$\therefore$$ $$\lambda_1 = \dfrac{h}{p_1}$$ and
$$\lambda_2 = \dfrac{h}{p_2}$$
$$p_1 = p_2$$ $$\implies$$ $$\lambda_1 = \lambda_2$$
Calculate the de Brogile wavelength of a photon whose linear momentum has a magnitude of $$3.3 \times 10^{-23}kgm/s$$.
Report Question
0%
0.0002 nm
0%
0.002 nm
0%
0.02 nm
0%
0.2 nm
0%
2 nm
Explanation
The
de Brogile wavelength of a photon is $$\lambda=\dfrac{h}{p}=\dfrac{6.6\times 10^{-34}}{3.3\times 10^{-23}}=2\times 10^{-11}=0.02\times 10^{-9} m=0.02 nm$$
Which of the following formula represents the energy of photon?
Report Question
0%
E=$$\dfrac{hc}{\lambda}$$
0%
E=$$hc \lambda$$
0%
E=$$\dfrac{h \lambda}{c}$$
0%
E=$$\dfrac{c \lambda}{h}$$
Explanation
The packets of light energy are called as photons.
Energy of a photon is given by $$E = h\nu = \dfrac{hc}{\lambda}$$
where
ν
and $$\lambda$$ are the frequency and wavelength of light, respectively.
Find the de-Broglie wavelength of an electron with kinetic energy of $$120\ eV$$.
Report Question
0%
$$95\ pm$$
0%
$$102\ pm$$
0%
$$112\ pm$$
0%
$$124\ pm$$
Explanation
Given : $$K = 120eV$$
Mass of electron $$m = 9.1\times 10^{-31}$$ kg
de-Broglie wavelength $$\lambda= \dfrac{h}{\sqrt{2mK}}$$
$$\therefore$$
$$\lambda= \dfrac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 120 \times 1.6\times 10^{-19}}}$$
Or $$\lambda = \dfrac{6.6\times 10^{-34}}{59.1\times 10^{-25}} = 0.112\times 10^{-9}$$ $$m = 112$$ $$pm$$
Laser is used in finding the distance between the moon and
the earth. The property of laser used in this process is,
Report Question
0%
Monochromaticity
0%
Non diverging nature
0%
High energy density
0%
Higher velocity
Explanation
Monochromaticity i.e. having one wavelength only is the property of laser which is used in the process to determine the distance between moon and earth.
If the momentum of an electron is increased by $${ P }_{ m }$$ then the de Broglie wavelength associated with it changes by $$0.5$$%. Then the initial momentum of the electron is
Report Question
0%
$$100{ P }_{ m }$$
0%
$$\dfrac { { P }_{ m } }{ 100 } $$
0%
$$200{ P }_{ m }$$
0%
$$\dfrac { { P }_{ m } }{ 200 } $$
Explanation
The De broglie wavelength $$\lambda$$ and momentum $$p$$ of a particle are related by ,
$$p=\dfrac{h}{\lambda}$$
as $$p\propto 1/\lambda$$ , therefore , with increase in momentum , wavelength will decrease .
Initially , $$p=\dfrac{h}{\lambda}$$ ................eq1
Now the wavelength is decreased by 0.5% ,
$$\lambda'=\lambda-\lambda\times0.5/100=199\lambda/200$$
hence , increased momentum ,
$$p+p_{m}=\dfrac{h}{\lambda'}$$
or
$$p+p_{m}=\dfrac{200h}{199\lambda}$$ ............eq2
dividing eq2 by eq1 , we get
$$\dfrac{p+p_{m}}{p}=\dfrac{200}{199}$$
or $$199p+200p_{m}=200p$$
or $$p=200p_{m}$$
Electrons used in an electron microscope are accelerated by a voltage of $$25\ kV$$. If the voltage is increased by $$100\ kV$$ then the de-Broglie wavelength associated with electrons would
Report Question
0%
increase by $$2$$ times
0%
decrease by $$2$$ times
0%
increase by $$4$$ times
0%
decrease by $$4$$ times
Explanation
de_broglie wavelength $$\lambda = \dfrac{h}{P}$$
where momentum of the particle $$P = \sqrt{2mK} = \sqrt{2meV}$$
$$\implies \ \lambda = \dfrac{h}{\sqrt{2meV}}$$
We get $$\lambda\propto \dfrac{1}{\sqrt{V}}$$
Given : $$V_i = 25kV$$ and $$V_2 = 100kV$$
So, voltage is increased by a factor of $$4$$.
Thus wavelength gets reduced by a factor of $$2$$.
The maximum kinetic energy of the photoelectrons depends only on
Report Question
0%
Potential
0%
Frequency
0%
Incident angle
0%
Pressure
Explanation
Maximum kinetic energy of photoelectrons $$K.E_{max} = h\nu - \phi$$
where $$\nu$$ is the frequency of incident photon and $$\phi$$ is the work function of metal
Hence maximum kinetic energy of photoelectrons depends only on the frequency of the incident photons.
The number of De-Broglie wavelengths contained in the second Bohr orbit of Hydrogen atom is
Report Question
0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
De-broglie wavelength of a particle is given as
$$\lambda=\dfrac{h}{mv}$$
According to the Bohr's postulate,
$$mvr=\dfrac{nh}{2\pi}$$
$$\implies 2\pi r=n\dfrac{h}{mv}=n\lambda$$
Hence there are $$n$$ De-Broglie wavelengths in $$n^{th}$$ Bohr orbit.
Hence correct answer is option B.
The de Broglie wavelength of an electron (mass = $$1 \times 10^{-30}$$ kg, charge = $$1.6 \times 10^{-1} C$$) with a kinetic energy of 200 eV is (Planck's constant = $$6.6 \times 10^{-34} J s$$)
Report Question
0%
$$9.60 \times 10^{-11} m$$
0%
$$8.25 \times 10^{-11} m$$
0%
$$6.25 \times 10^{-11} m$$
0%
$$5.00 \times 10^{-11} m$$
Explanation
Given : $$m=1\times 10^{-30}$$ kg $$e =1.6\times 10^{-19} C$$ $$h =6.6\times 10^{-34} Js$$
Kinetic energy $$K =200eV$$
de Broglie wavelength $$\lambda = \dfrac{h}{\sqrt{2mK}} =\dfrac{h}{\sqrt{2m (200) e}}$$
$$\therefore$$ $$\lambda = \dfrac{6.6\times 10^{-34}}{\sqrt{2(10^{-30}) (200) (1.6\times 10^{-19})}}$$
OR $$\lambda = \dfrac{6.6\times 10^{-34}}{8\times 10^{-24}} $$
$$\implies$$ $$\lambda = 8.25\times 10^{-11} m$$
What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 Volt?
Report Question
0%
12.27 $$\mathring{A}$$
0%
1.227 $$\mathring{A}$$
0%
0.1227 $$\mathring{A}$$
0%
0.001227 $$\mathring{A}$$
Explanation
Given : $$V =100$$ volts
de-Broglie wavelength of electron $$\lambda = \dfrac{h}{\sqrt{2meV}}$$
$$\therefore$$
$$\lambda = \dfrac{6.6\times 10^{-34}}{\sqrt{2(9.1\times 10^{-31})(1.6\times 10^{-19})(100)}}$$
Or
$$\lambda = \dfrac{6.6\times 10^{-34}}{5.4\times 10^{-10}}$$ m
$$\implies $$ $$\lambda = 1.227\times 10^{-10} m = 1.227$$ $$\mathring{A}$$
A source $$S_1$$ is producing, $$10^{15}$$ photons per second of wavelength 5000 $$A^0$$. Another source $$S_2$$ is producing $$1.02\times{10}^{15}$$ photon per second of wave length 5100 $$A^o$$, then ratio of power of Source $$S_2$$ and power of Source $$S_1$$
Report Question
0%
1.00
0%
1.02
0%
1.04
0%
2.00
Explanation
Power $$P = N\dfrac{hc}{\lambda}$$
where $$N$$ is the number of photons per second and $$\lambda$$ is the wavelength of light.
$$\implies \ P\propto \dfrac{N}{\lambda}$$
We get $$\dfrac{P_2}{P_1} = \dfrac{N_2\lambda_1}{N_1\lambda_2}$$
Given : $$N_1 = 10^{15}$$ $$N_2 = 1.02\times 10^{15}$$ $$\lambda_1 = 5000 \ A^o$$ $$\lambda_2 = 5100 \ A^o$$
$$\therefore$$ $$\dfrac{P_2}{P_1} = \dfrac{1.02\times 10^{15}\times 5000}{10^{15}\times 5100} = 1$$
An electron accelerated by a potential difference of V volt posses a de Broglie wave length . If the accelerating potential is increased by a factor of 4, the de-Broglie wavelength of the electron will be
Report Question
0%
remains unchanged
0%
becomes double
0%
becomes half
0%
becomes 4 times
Explanation
de-Broglie wavelength is given by $$\lambda = \dfrac{h}{p}$$
where, momentum of the particle $$p = \sqrt{2mK}$$
Kinetic energy of the particle $$K = eV$$
$$\implies \ \lambda = \dfrac{h}{\sqrt{2meV}}$$
We get $$\lambda\propto \dfrac{1}{\sqrt{V}}$$
Thus de-Broglie wavelength of the electron will become half if the accelerating potential is increased by a factor of $$4$$.
The work function of metals is in the range of $$2$$ eV to $$5$$ eV. Find which of the following wavelength of light cannot be used for photoelectric effect. (Consider, Planck constant $$=4\times10^{-15}$$eVs, velocity of light $$= 3\times 10^8m/s$$).
Report Question
0%
$$510nm$$
0%
$$650nm$$
0%
$$400nm$$
0%
$$570nm$$
Explanation
The minimum wavelength to cause photoelectric effect $$=\dfrac{hc}{E_{max}}$$
$$=\dfrac{4\times 10^{-15}\times 3\times 10^8}{5}m$$
$$=240nm$$
The maximum wavelength to cause photoelectric effect $$=\dfrac{hc}{E_{min}}$$
$$=\dfrac{4\times 10^{-15}\times 3\times 10^8}{2}m$$
$$=600nm$$
Hence light of wavelength $$650nm$$ is not acceptable.
A particle of mass 1kg is moving with a velocity of 1m/s. The de-broglie wavelength associated with it will be
Report Question
0%
h
0%
h/2
0%
h/4
0%
2h
Explanation
Given : $$v = 1m/s$$ $$m = 1 \ kg$$
de-Broglie wavelength $$\lambda = \dfrac{h}{mv}$$
$$\therefore$$ $$\lambda = \dfrac{h}{1\times 1} = h$$
The de-Broglie wavelength of a proton and alpha particle is same, the ratio of their velocities is :
Report Question
0%
1:2
0%
2:1
0%
1:4
0%
4:1
Explanation
de-Broglie wavelength, $$\lambda = \dfrac{h}{mv}$$
$$\implies \ v = \dfrac{h}{m\lambda}$$
We get $$\dfrac{v_{p}}{v_{\alpha}} = \dfrac{m_{\alpha}}{m_p}$$
Given : $$m_{\alpha} = 4m_p$$
$$\therefore$$ $$\dfrac{v_{p}}{v_{\alpha}} = \dfrac{4m_p}{m_p} = 4$$
Photo electric effect supports the quantum nature of light
because:
Report Question
0%
electric charge of photoelectrons is quantized.
0%
the maximum K.E. of photoelectrons depends only on the frequency of light and not on its intensity.
0%
even when the metal surface is faintly illuminated by light of the approximate wavelength, the photo electrons leave the surface immediately.
0%
none of these
Explanation
Photo electric effect supports the quantum nature of light because even when the metal surface is faintly illuminated by light of the approximate wavelength, the photo electrons leave the surface immediately.
The energy of photon of wavelength $$\lambda$$ is
[h = Planck's constant, c = speed of light in vacuum]
Report Question
0%
$$hc \lambda$$
0%
$$\dfrac{h\lambda}{c}$$
0%
$$\dfrac {\lambda}{hc}$$
0%
$$\dfrac {hc}{\lambda}$$
Explanation
The energy of a photon is directly proportional to its frequency with proportionality constant being the Plank's constant $$h$$.
Hence $$E=h\nu=h\dfrac{c}{\lambda}$$(Since $$c=\lambda\nu$$)
$$\implies E=\dfrac{hc}{\lambda}$$
If the kinetic energy of the moving particle is $$E$$, then the de Broglie wavelength is
Report Question
0%
$$\lambda = \dfrac {h}{\sqrt {2mE}}$$
0%
$$\lambda = \dfrac {\sqrt {2mE}}{h}$$
0%
$$\lambda = h \sqrt {2mE}$$
0%
$$\lambda = \dfrac {h}{E\sqrt {2m}}$$
Explanation
De Broglie wavelength of the particle, $$\lambda = \dfrac{h}{p}$$ where $$p$$ is the momentum of the particle
Kinetic energy of particle, $$E = \dfrac{1}{2}mv^2 = \dfrac{p^2}{2m}$$ $$(\because p = mv)$$
Thus, we get: $$p = \sqrt{2mE}$$
$$\implies$$
$$\lambda = \dfrac{h}{\sqrt{2mE}}$$
Blue light has a frequency of approximately $$6.0 \times 10^{14}$$ hertz. A photon of blue light will have a
Report Question
0%
$$4.0 \times 10^ {-19} J$$
0%
$$1.1 \times 10^ {-48} J$$
0%
$$ 6.0 \times 10^ {-34} J$$
0%
none
Explanation
Frequency of the blue light $$\nu = 6\times 10^{14} \ Hz$$
Energy of photon $$E = 6.626\times 10^{-34}\times 6\times 10^{14} = 4\times 10^{-19} \ J$$
Correct answer is option A.
A particle $$A$$ of mass $$m$$ and initial velocity $$v$$ collides with a particle $$B$$ of mass $$\dfrac{m}{2}$$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $${\lambda}_A$$ to $${\lambda}_B$$ after the collision is :
Report Question
0%
$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{1}{2}$$
0%
$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{1}{3}$$
0%
$$\dfrac{{\lambda}_A}{{\lambda}_B}=2$$
0%
$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{2}{3}$$
Explanation
From conservation of linear momentum, $$mv= mv_B + \dfrac{m}{2} v_B$$
$$v= v_A+ \dfrac{v_B}{2}$$
Also $$ v^2 = v_a^2 + \dfrac{v_B^2}{2}$$
$${(v_A+ \dfrac{v_B}{2})}^2= v_A^2 + {(\dfrac{v_B}{2})}^2$$
$$v_Av_B = \dfrac{v_B}{4}$$
$$v_B= 4 v_A$$ and $$m_B= \dfrac{m_A}{2}$$
$$P_A=m_Av_A$$
$$P_B= \dfrac{m_A}{2} \times 4 v_A= 2 P_A$$
$$\dfrac{\lambda_A}{\lambda_B}= \dfrac{P_B}{P_A}= \dfrac{2 P_A}{P_A}= 2$$
The kinetic energy of an electron get tripled then the de-Broglie wavelength associated with electron changes by a factor of
Report Question
0%
$$\dfrac { 1 }{ 3 } $$
0%
$$\sqrt { 3 } $$
0%
$$\dfrac { 1 }{ \sqrt { 3 } } $$
0%
$$3$$
Explanation
de-Broglie wavelength of an electron
$$\lambda =\dfrac { h }{ \sqrt { 2mK } }$$
Or $$ \lambda \propto \dfrac { 1 }{ \sqrt { K } }$$
where $$K$$ is the kinetic energy of the electron and $$m$$ is the mass.
$$ \therefore \dfrac { { \lambda }^{ \prime } }{ \lambda } =\dfrac { 1 }{ \sqrt { 3K } } .\dfrac { \sqrt { K } }{ 1 } =\dfrac { 1 }{ \sqrt { 3 } }$$
Or $$ { \lambda }^{ \prime }=\dfrac { \lambda }{ \sqrt { 3 } }$$
i.e., de-Broglie wavelength will decrease by a factor of $$ \dfrac { 1 }{ \sqrt { 3 } } $$.
$$A$$ and $$B$$ are two metals with threshold frequencies $$1.8\times { 10 }^{ 14 }Hz$$ and $$2.2\times { 10 }^{ 4 }Hz$$. Two identical photons of energy $$0.825 eV$$ each are incident on them. Then photoelectrons are emitted by (Take $$h=6.6\times { 10 }^{ -34 }J-s$$)
Report Question
0%
$$B$$ alone
0%
$$A$$ alone
0%
Neither $$A$$ nor $$B$$
0%
Both $$A$$ and $$B$$
Explanation
Threshold energy of $$A$$: $${ E }_{ A }=h{ v }_{ A }$$
$$=6.6\times { 10 }^{ -34 }\times 1.8\times { 10 }^{ 14 }$$
$$=11.88\times { 10 }^{ -20 }J$$
$$=\dfrac { 11.88\times { 10 }^{ -20 } }{ 1.6\times { 10 }^{ -19 } } eV=0.74eV$$
Similarly, $${ E }_{ B }=0.91eV$$
As the incident photons have energy greater than $${ E }_{ A }$$ but less than $${E}_{B}$$
So, photoelectrons will be emitted from metal $$A$$ only.
A particle is dropped from a height '$$H$$'. The de Broglie wavelength of the particle depends on height as
Report Question
0%
$$H$$
0%
$${ H }^{ -{ 1 }/{ 2 } }$$
0%
$${ H }^{ 0 }$$
0%
$${ H }^{ { 1 }/{ 2 } }$$
Explanation
Let the velocity of particle just before it reaches the ground be $$V$$.
Using energy conservation of the particle, we get
$$\dfrac{1}{2}mV^2 = mgH$$
$$\implies$$ $$V = \sqrt{2gH}$$
de Broglie wavelength of the particle $$\lambda = \dfrac{h}{mV}$$
$$\therefore$$ $$\lambda = \dfrac{h}{m \sqrt{2gH}}$$
$$\implies$$ $$\lambda \propto H^{-1/2}$$
A stream of photons having energy 3 eV each impinges on a potassium surface. The work function of potassium is 2.3 eV. The emerging photo-electrons are slowed down by a copper plate placed 5 mm away. If the potential difference between the two metal plates is 1 V, the maximum distance the electrons can move away from the potassium surface before being turned back is.
Report Question
0%
3.5 mm
0%
1.5 mm
0%
2.5 mm
0%
5.0 mm
Explanation
Given, the Work function of Potassium $$w$$ is $$2.3 eV$$
Energy of incident photon beam is $$3eV$$
Thus, the maximum Kinetic Energy of the photo electrons is $$KE = 3-2.3=0.7eV$$
At the point of returning back, the entire Kinetic Energy is converted into Electrostatic Potential Energy (Statement 1)
The Electric field between the plates is given by $$E = \frac{\Delta V}{\Delta x} = \frac{1V}{5 mm} = 0.2 Vmm^{-1}$$
Let the distance travelled by the electron be $$x$$.
Increase in Electrostatic Potential energy = $$E \times x$$
Thus, (from statement 1)
$$e\times 0.2 Vmm^{-1} \times x = 0.7eV \Rightarrow x = 3.5mm$$
For light of wavelength $$\lambda$$ in nanometer, the photon energy $$hf$$ in electron-volt is
Report Question
0%
$$\dfrac{1240}{\lambda}$$
0%
$$\dfrac{1200}{\lambda}$$
0%
$$\dfrac{\lambda}{1240}$$
0%
$$\dfrac{1360}{\lambda}$$
Explanation
Photon energy is given by:
$$E = \cfrac{hc}{\lambda}$$
$$ = \cfrac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{\lambda \times 10^{-9}}\ J$$
$$ \approx \cfrac{2 \times 10^{-16}}{\lambda} \times 6.2 \times 10^{18}\ eV$$
$$ = \cfrac{1240}{\lambda}\ eV$$
A $$160$$watt light source is radiating light of wavelength $$6200\overset{o}{A}$$ uniformly in all directions. The photon flux at a distance of $$1.8$$m is of the order of (Plank's constant $$6.63\times 10^{-34}J-s$$).
Report Question
0%
$$10^2m^{-2}s^{-1}$$
0%
$$10^{12}m^{-2}s^{-1}$$
0%
$$10^{19}m^{-2}s^{-1}$$
0%
$$10^{25}m^{-2}s^{-1}$$
Explanation
Intensity of light $$I=\dfrac{160}{4\pi (1.8)^2}$$
Photon flux =$$\dfrac{I}{hc/\lambda}=1.22\times 10^{19}$$
The de Broglie wavelength of an electron is $$0.4\times { 10 }^{ -10 }m$$ when its kinetic energy is $$1.0keV$$. Its wavelength will be $$1.0\times { 10 }^{ -10 }m$$, when its kinetic energy is
Report Question
0%
$$0.2keV$$
0%
$$0.8keV$$
0%
$$0.63keV$$
0%
$$0.16keV$$
Explanation
$$\lambda =\cfrac { h }{ p } ,\lambda =\cfrac { h }{ \sqrt { 2mk } } $$
so, $$\lambda \propto \cfrac { 1 }{ \sqrt { k } } $$
$$\Rightarrow \cfrac { 0.4\times { 10 }^{ -10 } }{ 1.0\times { 10 }^{ -10 } } =\cfrac { \sqrt { k } }{ \sqrt { 1 } } \Rightarrow k=0.16keV$$
Ultraviolet light of wavelength $$300nm$$ and intensity $$1.0W/{m}^{2}$$ falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photo electrons emitted from an area of $$1.0{cm}^{2}$$ of the surface is nearly
Report Question
0%
$$19.61\times { 10 }^{ 12 }{ s }^{ -1 }$$
0%
$$4.12\times { 10 }^{ 12 }{ s }^{ -1 }$$
0%
$$1.51\times { 10 }^{ 12}{ s }^{ -1 }$$
0%
$$2.13\times { 10 }^{ 12}{ s }^{ -1 }$$
Explanation
Energy incident over $$1m^2:$$ $$=1\times10^{-4}J$$
Energy required to produce photoelectron
$$=1\times10^{-4}\times10^{-2}=10^{-6}J$$
Number of photoelectrons ejected $$=$$ Number of photons which can produce photoelectrons $$=$$ (energy required for producing electron)/(energy of photon) $$= \dfrac{10^{-6}}{h(c/\lambda)}=\dfrac{10^{-6}\times300\times10^{-9}}{6.6\times10^{-34}\times3\times10^8}=1.51\times10^{12}s^{-1}$$
Photons of energy $$7 e V$$ are incident on two metals A and B with work functions $$6 eV$$ and $$3 eV$$ respectively. The minimum de Broglie wavelengths of the emitted photoelectrons with maximum energies are $$\lambda_A$$ and $$\lambda_B$$ , respectively where $$\lambda_A / \lambda_B $$ is nearly :
Report Question
0%
$$0.5$$
0%
$$1.4$$
0%
$$4.0$$
0%
$$2.0$$
Explanation
Kinetic energy of photo electrons $$K = E_{photons} - \phi$$
where $$\phi$$ is the work function of the metal and $$E_{photon}$$ is the energy of the photon.
For metal A :
$$K_A = 7-6 =1$$ eV
For metal B :
$$K_B = 7-3 =4$$ eV
de Broglie wavelength $$\lambda = \dfrac{h}{\sqrt{2mK}}$$
$$\implies$$ $$\dfrac{\lambda_A}{\lambda_B} = \sqrt{\dfrac{K_B}{K_A}}$$
Or
$$\dfrac{\lambda_A}{\lambda_B} = \sqrt{\dfrac{4}{1}} = 2.0$$
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is :
Report Question
0%
Zero
0%
Less than that of a proton
0%
More than that of a proton
0%
Equal to that of a proton
Explanation
de Broglie wavelength $$\lambda = \dfrac{h}{\sqrt{2mK}}$$
We get $$K = \dfrac{h^2}{2m \lambda^2}$$
$$\implies$$ $$K\propto \dfrac{1}{m}$$
We know that $$m_p > m_e$$
$$\therefore$$ $$\dfrac{K_e}{K_p} = \dfrac{m_p}{m_e} >1$$
$$\implies$$ $$K_e > K_p$$
The de-Broglie wavelength '$$\lambda$$' of a particle
Report Question
0%
Is proportional to mass
0%
Is proportional to impulse
0%
Is inversely proportional to impulse
0%
Does not depend on impulse
Explanation
De Broglie wavelength of the particle $$\lambda = \dfrac{h}{mv}$$
We define impulse as the change in the momentum of the particle i.e. $$I = mv$$
We get $$\lambda = \dfrac{h}{I}$$
$$\implies$$ $$\lambda \propto \dfrac{1}{I}$$
A proton and an alpha particle are subjected to same potential difference $$V$$. Their de-Broglies wavelengths $$\lambda_{p}, \lambda_{\alpha}$$ will be in the ratio.
Report Question
0%
$$2 : 1$$
0%
$$2\sqrt {2} : 1$$
0%
$$4 : 1$$
0%
$$1 : 2$$
Explanation
Let the mass of alpha be $$4m$$ and charge be $$2e$$
therefore mass of proton is $$m$$ and charge is $$e$$
Energy of proton $$E_P=e\times V$$
So momentum is $$P_P=\sqrt{2Em}=\sqrt{2eVm}$$
Similar for alpha $$P_{\alpha}=\sqrt{2\times2e\times V\times 4m}$$
$$\lambda_{p}=\dfrac{h}{P_p}$$
$$\lambda_{\alpha}=\dfrac{h}{P_{\alpha}}$$
$$\dfrac{\lambda_{\alpha}}{\lambda_{P}}=2\sqrt{2}$$
Two identical metal plates show photoelectric effect by a light of wavelength $$\lambda _1$$ on plate 1 and $$\lambda _2$$ on plate 2 (where $$\lambda _1=2 \lambda_2$$ ). The maximum kinetic energy will be :
Report Question
0%
$$2K_2=K_1$$
0%
$$K_1 < K_2/2$$
0%
$$K_1>K_2/2$$
0%
$$2K-1=K_2$$
Explanation
Maximum kinetic energy $$K = \dfrac{hc}{\lambda} - \phi$$
where $$\lambda$$ is the wavelength of incident light and $$\phi$$ is the work function of the metal.
For first case : $$\lambda = \lambda_1 = 2\lambda_2$$
$$K_1 = \dfrac{hc}{2\lambda_2} - \phi$$ ....(1)
For second case : $$\lambda = \lambda_2$$
$$K_2 = \dfrac{hc}{\lambda_2} - \phi$$
$$\dfrac{K_2}{2} = \dfrac{hc}{2\lambda_2} - \dfrac{\phi}{2}$$
.....(2)
From (2) - (1), we get
$$\dfrac{K_2}{2} - K_1 = \dfrac{\phi}{2} $$
$$\implies \dfrac{K_2}{2} > K_1 $$
A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :
Report Question
0%
$$2\sqrt {2}$$
0%
$$\dfrac {1}{2\sqrt {2}}$$
0%
$$2$$
0%
$$\sqrt {2}$$
Explanation
Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity $$v$$ is given by
$$\lambda = \dfrac {h}{mv} $$ $$ (\because p = mv)$$
de-Broglie wavelength of a proton of mass $$m_{1}$$ and kinetic energy $$k$$ is given by
$$\lambda_{1} = \dfrac {h}{\sqrt {2m_{1}k}}$$ $$(\because p = \sqrt {2mk})$$
$$\lambda_1= \dfrac {h}{\sqrt {2m_{1}qV}} .... (i)$$ $$ [\because k = qV]$$
For an alpha particle mass $$m_{2}$$ carrying charge $$q_{0}$$ is accelerated through potential $$V$$, then
$$\lambda_{2} = \dfrac {h}{\sqrt {2m_{2}q_{0}V}}$$
$$\because$$ For $$\alpha - particle\ \ (^{4}_{2}He)$$ : $$ q_{0} = 2q$$ and $$m_{2} = 4m_{1}$$
$$\therefore \lambda_{2} = \dfrac {h}{\sqrt {2\times 4m_{1}\times 2q\times V}} .... (ii)$$
The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get
$$\dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {h}{\sqrt {2m_{1}qV}} \times \dfrac {\sqrt {2\times m_{1} \times 4\times 2qV}}{h} = \dfrac {4}{\sqrt {2}} \times \dfrac {\sqrt {2}}{\sqrt {2}}$$
We get $$\dfrac{\lambda_1}{\lambda_2}= 2\sqrt {2}$$
A gamma ray photon creates an electron, positron pair. If the rest mass energy of an electron is 0.5 MeV and the total KE of the electron, positron pair is 0.78 MeV, then the energy of the gamma ray photon must be :
Report Question
0%
0.78 MeV
0%
1.78 MeV
0%
1.28 MeV
0%
0.28 MeV
Explanation
Energy of $$\gamma - ray$$ photon $$=0.5+0.5+0.78$$
$$=1.78 \, MeV$$
The ionastion energy of the electron in the hydrogen atom in its ground state is $$13.6 eV$$. The atoms are excited to higher energy levels to emit radiations of $$6$$ wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between :
Report Question
0%
$$n = 3$$ to $$n = 1$$ states
0%
$$n = 4$$ to $$n = 3$$ states
0%
$$n = 3$$ to $$n = 2$$ states
0%
$$n = 2$$ to $$n = 1$$ states
Explanation
Number of wavelengths $$ = \dfrac {n(n-1)}{2} $$
where, n = number of orbit from which transition takes place.
$$ \therefore 6 = \dfrac {n(n-1)}{2} $$
$$n = 4$$
$$ \therefore $$ The wavelength of emitted radiations will be maximum for transition $$n = 4$$ to $$n = 3.$$
A radio transmitter operates at a frequency of 880 kHz and power of 10 kW. The number of photons emitted per second is :
Report Question
0%
$$13.27 \times 10^4$$
0%
$$13.27 \times 10^{34}$$
0%
$$1327 \times 10^{34}$$
0%
$$1.71 \times 10^{31}$$
Explanation
Power of transmitter $$P =10 \ kW = 10 \times 10^3 \ W$$
Frequency of the transmitter $$\nu = 880 \ kHz = 880\times 10^3 \ Hz$$
Number of photons emitted per second
$$N= \dfrac{P}{h\nu} = \dfrac{10 \times 10^3}{6.6 \times 10^{-34} \times 880 \times 10^3}$$
$$N= 1.71 \times 10^{31}$$ photons emitted per second.
A photon and an electron possesses same de-Broglie wavelength. Given that C$$=$$ speed of light and v$$=$$ speed of electron, which of the following relation is correct? (Here, $$E_e=K.E$$ of electron, $$E_{Ph}=K.E$$ of photon, $$P_e=$$ momentum of electron, $$P_{ph}=$$ momentum of photon).
Report Question
0%
$$\displaystyle\frac{P_e}{P_{Ph}}=\frac{C}{2v}$$
0%
$$\displaystyle\frac{E_e}{E_{Ph}}=\frac{C}{2v}$$
0%
$$\displaystyle\frac{E_{ph}}{E_e}=\frac{2c}{v}$$
0%
$$\displaystyle\frac{P_e}{P_{Ph}}=\frac{2c}{v}$$
Explanation
We have, $$\displaystyle \lambda_{Ph}=\frac{h}{P_{Ph}}$$ and $$\displaystyle \lambda_e=\frac{h}{P_e}$$
Given, $$\displaystyle\lambda_{Ph}=\lambda_e$$
$$\therefore$$ We get, $$P_{Ph}=P_e$$
$$\displaystyle\frac{h}{\lambda_{Ph}}=mv$$
$$\therefore \displaystyle\frac{hc}{\lambda_{Ph}}=mcv=\frac{1}{2}mv^2\left(\displaystyle\frac{2c}{v}\right)$$
or $$\displaystyle\frac{E_{Ph}}{E_e}=\frac{2c}{v}$$.
A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to
Report Question
0%
$$H$$
0%
$$H^{1/2}$$
0%
$$H^{0}$$
0%
$$H^{-1/2}$$
Explanation
Velocity acquired by a particle while talling from a height m is, $$\displaystyle v= \sqrt{2gH}$$
Thus, as, $$\displaystyle \lambda = \frac{h}{mv} = \frac{h}{m \sqrt{2gH}}$$
or $$\displaystyle \lambda \propto H^{-1/2}$$
A photon of energy 10.2 eV corresponds to light of wavelength $$\lambda_0$$. Due to an electron transition from x =2 to x = 1 in a hydrogen atom, light of wavelength A is emitted. If we take into account the recoil of the atom when the photon is emitted, then :
Report Question
0%
$$\lambda <\lambda _0$$
0%
$$\lambda >\lambda _0$$
0%
$$\lambda = \lambda _0$$
0%
None of these
Explanation
Without recoil of atom
$$\dfrac{hc}{\lambda_0}=10.2 eV$$
With recoil of atom
$$\underset{\cdot }{\rightarrow} =\underset{m}{\underset{\cdot }{\leftarrow }}+m\rightarrow \dfrac{h}{\lambda }$$
$$\dfrac{hc}{\lambda }+\dfrac{1}{2}mv^2=10.2$$
$$\dfrac{hc}{\lambda }<\dfrac{hc}{\lambda _0}$$
$$\therefore \lambda _0<\lambda $$
A photon of energy $$hv$$ and momentum $$hv/c$$ collides with an electron at rest. After the collision, the scattered electron and the scattered photon each make an angle of $$45^{\circ}$$ with the initial direction of motion. The ratio of frequency of scattered and incident photon is :
Report Question
0%
$$\sqrt {2}$$
0%
$$\sqrt {2} - 1$$
0%
$$2$$
0%
$$1/\sqrt {2}$$
Explanation
This is an example of Compton Scattering, which is scattering of a photon by an electron. It results in a decrease in energy (thus decrease in frequency) of the photon.
Let the frequency of scattered photon be $$\nu^{'}$$
By conservation of momentum in vertical direction, the momentum of scattered electron and photon are same and equal to $$h\nu^{'}/c$$.
By conservation of linear momentum in horizontal direction,
$$h\nu/c = 2\times h\nu^{'}cos45^0/c\implies \nu^{'}/\nu = 1/\sqrt{2}$$
So option D is correct.
Out of a photon and an electron, the equation $$E = Pc$$, is valid for
Report Question
0%
both
0%
neither
0%
photon only
0%
electron only
Explanation
Energy is given by
$$E=\dfrac { { m }_{ 0 }{ c }^{ 2 } }{ \sqrt { 1-\dfrac { { v }^{ 2 } }{ { c }^{ 2 } } } } $$
$$\Rightarrow { E }^{ 2 }=\dfrac { { m }_{ 0 }^{ 2 }{ c }^{ 6 } }{ { c }^{ 2 }-{ v }^{ 2 } } $$
Momentum $$P$$ is given by
$$P=\dfrac { { m }_{ 0 }v }{ \sqrt { 1-\left( \dfrac { { v }^{ 2 } }{ { c }^{ 2 } } \right) } } $$
$${ P }^{ 2 }{ c }^{ 2 }=\dfrac { { m }_{ 0 }^{ 2 }{ c }^{ 4 }{ v }^{ 2 } }{ { c }^{ 2 }-{ v }^{ 2 } } $$
$$\therefore { E }^{ 2 }-{ P }^{ 2 }{ c }^{ 2 }={ m }_{ 0 }^{ 2 }{ c }^{ 4 }$$
$$\Rightarrow { E }^{ 2 }={ P }^{ 2 }{ c }^{ 2 }+{ m }_{ 0 }^{ 2 }{ c }^{ 4 }$$
For photon, rest mass
$${ m }_{ 0 }=0$$, so $$E=Pc$$
For electron, $${ m }_{ 0 }\neq 0$$, so $$E\neq Pc$$
The de Broglie wavelength associated with a ball of mass 150 g travelling at 30 m $$s^{-1}$$ is
Report Question
0%
$$1.47 \, \times \, 10^{-34}m$$
0%
$$1.47 \, \times \, 10^{-16}m$$
0%
$$1.47 \, \times \, 10^{-19}m$$
0%
$$1.47 \, \times \, 10^{-31}m$$
Explanation
Mass of the bail, m = 150 g = 0.15 kg.
speed of the ball, v = 30 $$m \, s^{-1}$$
Momentum, $$p = mv = 0.15 \, \times \, 30 \, = \, 4.5 kg \, m \, s^{-1}$$
de Broglie wavelength,
$$\lambda = \dfrac{h}{p} \, = \, \dfrac{6.63 \, \times \, 10^{-34}}{4.5} \, = \, 1.47 \, \times \, 10^{-34} m$$
If levels $$1$$ and $$2$$ are separated by an energy $${ E }_{ 2 }-{ E }_{ 1 }$$, such that the corresponding transition frequency falls in the middle of the visible range, calculate the ratio of the populations of two levels in the thermal equilibrium at room temperature.
Report Question
0%
$$1.1577\times { 10 }^{ -38 }$$
0%
$$2.9\times { 10 }^{ -35 }$$
0%
$$2.168\times { 10 }^{ -36 }$$
0%
$$1.96\times { 10 }^{ -20 }$$
Explanation
At thermal equilibrium, the ratio $$\dfrac { { N }_{ 2 } }{ { N }_{ 1 } } $$ is given as
$$\dfrac { { N }_{ 2 } }{ { N }_{ 1 } } =exp \left( -\dfrac { { E }_{ 2 }-{ E }_{ 1 } }{ kT } \right) $$
The middle of the visible range is taken at $$\lambda =550 nm$$
$$\Rightarrow { E }_{ 2 }-{ E }_{ 1 }=\dfrac { hc }{ \lambda } =3.16\times { 10 }^{ -19 }J$$
$$\Rightarrow \dfrac { { N }_{ 2 } }{ { N }_{ 1 } } =exp \left( \dfrac { -3.16\times { 10 }^{ -19 }J }{ \left( 1.38\times { 10 }^{ -23 }{ 1 }/{ k } \right) \cdot \left( 300k \right) } \right) $$
$$\Rightarrow \dfrac { { N }_{ 2 } }{ { N }_{ 1 } } =1.1577\times { 10 }^{ -38 }$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Medical Physics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
