instrument for measuring light intensity

MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 12

The kinetic energy of electron and proton is

$10^{-32}\ J$ . Then the relation between their de-Broglie wavelength is

0%
$\lambda_p < \lambda_e$
0%
$\lambda_p > \lambda_e$
0%
$\lambda_p = \lambda_e$
0%
$\lambda_p =2 \lambda_e$

Explanation

According to the question,

By using

$\lambda =\dfrac {h}{\sqrt{2mE}}\ E=10^{-32}\ J$ = Constant for both
particles. Hence

$\lambda \propto \dfrac {1}{\sqrt m}$ Since

$m_p > m_e$ so

$\lambda_p < \lambda_e$ .

AN important spectral emission line has a wavelength of

$21\ cm$ . The corresponding photon energy is

0%
$5.9\times 10^{-4}eV$
0%
$5.9\times 10^{-6}eV$
0%
$5.9\times 10^{-8}eV$
0%
$11.8\times 10^{-6}eV$

Explanation

As we know,

$E=\dfrac{hc}{\lambda}=\dfrac{3\times 10^8 \times 6.62 \times 10^{-34}}{0.21 \times 1.6 \times 10^{-19}}=5.9\times 10^{-6}eV$

Which of the following is true for photon

0%
$E=\dfrac {hc}{\lambda}$
0%
$E=\dfrac {1}{2}mu^2$
0%
$p=\dfrac {E}{2v}$
0%
$E=\dfrac {1}{2}mc^2$

Explanation

For photon,

$E=\dfrac{hc}{\lambda}$ .

The de-Broglie wavelength is proportional to

0%
$\lambda \propto \dfrac 1v$
0%
$\lambda \propto \dfrac 1m$
0%
$\lambda \propto \dfrac 1p$
0%
$\lambda \propto p$

Explanation

As we know,

$\lambda =\dfrac hp$

$\Rightarrow \lambda \propto \dfrac 1p$

The spectrum of radiation

$1.0\times 10^{14}Hz$ is the infrared region. The energy of one photon of this in joules will be

0%
$6.62\times 10^{-48}$
0%
$6.62\times 10^{-20}$
0%
$\dfrac {6.62}{3}\times 10^{-28}$
0%
$3\times 6.62\times 10^{-28}$

Explanation

As we know,

$E=hv=6.64\times 10^{-34}\times 1.0\times 10^{14}=6.62\times 10^{-20}J$

A laser is a coherent source because it contains

0%
Many wavelengths
0%
Unconditinated wave of a particular wavelength
0%
Coordinated wave of many wavelengths
0%
Coordinated wave of particular wavelengths

Explanation

LASER is the short form of Light Amplification by Stimulated Emission of Radiation. Laser beam is intense, monochromatic (that is of the single wavelength) , collimated and highly coherent.Hence , correct option is

$(D)$ .

The minimum wavelength of photons is

$5000 \overset {o}{A}$ , its energy will be

0%
$2.5\ eV$
0%
$50\ eV$
0%
$5.48\ eV$
0%
$7.48\ eV$

Explanation

As we know,

$E=\dfrac {12375}{\lambda}=\dfrac {12375}{5000}=2.47\ eV\approx 2.5\ eV$

If the energy of a photon correspoiding to a wavelength of

$6000\overset {o}{A}$ is

$3.32\times 10^{-19}J$ , the photon energy for a wavelength of

$4000\overset {o}{A}$ will be

0%
$1.4\ eV$
0%
$4.9\ eV$
0%
$3.1\ eV$
0%
$1.6\ eV$

Explanation

As we know,

$E=\dfrac {hc}{\lambda}\Rightarrow \dfrac {E_1}{E_2}=\dfrac {\lambda_1}{\lambda_2}\Rightarrow \dfrac {3.32\times 10^{-19}}{E_2}=\dfrac {4000}{6000}$

$\Rightarrow E_2=4.98\times 10^{-19}J=3.1\ eV$

de-Broglie wavelength of a body of mass

$m$ and kinetic energy

$E$ is given by

0%
$\lambda =\dfrac {h}{mE}$
0%
$\lambda =\dfrac {\sqrt{2mE}}{h}$
0%
$\lambda =\dfrac {h}{2mE}$
0%
$\lambda =\dfrac {h}{\sqrt{2mE}}$

Explanation

de Broglie wavelength associated with the particle

$\lambda =\dfrac hp$

where h is the Planck's constant

now we know thwt ,

$E=\dfrac12 mv^2$ and

$p=mv$

by mutiplying by m on both sides we get,

$p=\sqrt{2mE}$

hence,

$\lambda =\dfrac h{\sqrt{2mE}}$

A proton and an

$\alpha$ particle are acclerated through a potential difference of

$100\ V$ . The ratio of the wavelength associated with the proton to that associated with an

$\alpha$ - particle is

0%
$\sqrt 2:1$
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$2:1$
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$2\sqrt 2:1$
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$\dfrac{1}{2\sqrt 2}:1$

Explanation

As we know,

$\lambda =\dfrac{h}{\sqrt{2mQV}}\Rightarrow \propto \dfrac{1}{\sqrt{mQ}}\Rightarrow \dfrac{\lambda_p}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha}Q_{\alpha}}{m_p Q_p}}$

$=\sqrt{\dfrac{4m_p\times 2Q_p}{m_p\times Q_p}}=2\sqrt 2$

The log-log graph between the energy

$E$ of an electron and its de-Broglie wavelength

$\lambda$ will be

Explanation

As we know,

$\lambda =\dfrac h{\sqrt{2mE}}$

taking log on both sides, we get,

$log\lambda=logh-\dfrac 12 log 2m -\dfrac 12logE$

hence, graph will be a straight line with negative slope.

option C is correct.

For moving ball of cricket, the correct statement about de-Broglie wavelength is

0%
It is not applicable for such big particle
0%
$\dfrac{h}{\sqrt{2mE}}$
0%
$\sqrt{\dfrac{h}{2mE}}$
0%
$\dfrac{h}{2mE}$

Explanation

de Broglie wavelength associated with the particle

$\lambda =\dfrac hp$

where h is the Planck's constant

now we know thwt ,

$E=\dfrac12 mv^2$ and

$p=mv$

by mutiplying by m on both sides we get,

$p=\sqrt{2mE}$

hence,

$\lambda =\dfrac h{\sqrt{2mE}}$

An electron is moving through a field. It is moving (i) opposite an electric field (ii) perpendicular to a magnetic field as shown. For each situation the de-Broglie wave length of electron

0%
Increasing, Increasing
0%
Increasing,decreasing
0%
Decreasing, same
0%
same,same

Explanation

$\lambda=\dfrac{h}{mv}$ . Since

$v$ increasing in case (i), but it is not changing in case (ii). Hence, in the first case de-Broglie wavelength will change, but it second case, it remain the same

Find out the de-Broglie wavelength related to an electron of kinetic energy

$10$ eV:

0%
$10 \mathring A$
0%
$1227 \mathring A$
0%
$0.10 \mathring A$
0%
$3.9 \mathring A$

Explanation

$\lambda = \dfrac{h}{\sqrt{2mE}}$

$\lambda = \dfrac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 10 \times 1.6 \times 10^{-19}}}$

$= 3.9 \mathring A$

The output from a LASER is monochromatic. It means that it is

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Directional
0%
Polarised
0%
Narrow beam
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Single frequency

Given below are two statements:
Statement I: Two photons having equal linear momenta have equal wavelengths.
Statement- II: If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.

0%
Both Statement I and Statement II are true.
0%
Statement I is false but Statement II is true
0%
Both Statement I and Statement II are false
0%
Statement I is true but Statement II is false

Explanation

If linear momentum are equal then wavelength also equal

$p = \dfrac{h}{\lambda} , E = \dfrac{hc}{\lambda}$

On decreasing wavelength, momentum and energy of photon increases.

The wavelength of the de-Broglie wave associated with a thermal neutron of mass

$m$ at absolute temperature

$T$ is given by:

(

$k$ is the Boltzmann constant)

0%
$\displaystyle \frac{h}{\sqrt{mkT}}$
0%
$\displaystyle \frac{h}{\sqrt{2mkT}}$
0%
$\displaystyle \frac{h}{\sqrt{3mkT}}$
0%
$\displaystyle \frac{h}{2\sqrt{mkT}}$

Explanation

The kinetic energy of the thermal neutron of mass m at absolute temperature T is given by

$\dfrac {mv^2}{2}=kT$

$\Rightarrow p=mv=\sqrt {2mkT}$

The wavelength of the deBroglie wave associated with the thermal neutron is

$\lambda=h/p=\dfrac{h}{\sqrt {2mkT}}$ .

What should be the approximate K.E. of an electron so that its de -Broglie wavelength is equal to the wavelength of x -ray of maximum energy produced in an x -ray tube operating at 24,800 V?

$(h = 6.6 /\times 10^{-34}J - sec$ , mass of electron

$m =9.1 \times 10^{-31} kg)$ (give answer in

$10^2 eV).$

0%
$3$
0%
$4$
0%
$6$
0%
$9$

Photoelectrons emitted from a photo sensitive metal of work function

$1eV$ describe a circle of radius 0.1 cm in a magnetic field of induction

$10^{-3}$ Tesla. The energy of the incident photons is (mass of electron

$= 9 \times 10^{-31}$ kg)

0%
$1.17 eV$
0%
$2.9 eV$
0%
$0.9 eV$
0%
$0.81 eV$

Explanation

$=\dfrac{mv}{r}=Bq$

$V=\dfrac{Bqr}{m}=Bq$

Velocity of electrons rotating i the magnetic field

$=\dfrac{Bqn}{m}$

$=\dfrac{10^{-3}\times1.6\times 10^{-19}\times 0.1\times10^{-2}}{9.1 \times10^{-31}}$ m/s

So,

energy of electron

$=\dfrac{1}{2}mv^{2}=\dfrac{1}{2}\times \dfrac{B^{2}e^{2}n^{2}}{m^{2}}$

$=\dfrac{10^{-6}\times \left ( 1.6 \times10^{-19} \right )\times10^{-6}}{9.1\times10^{-31}}$ eV

$=\dfrac{1.6}{9}eV=0.17eV$

So,

$=$ energy of incident photon

$= (1+0.17) eV$

$= 1.17 eV$

A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is :

(take the proton mass,

$m_p = (5/3) \times 10^{-27}kg$ and

$h/e = 4.2\times 10^{-15}J.s/C; \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 m/F; 1 fm = 10^{-15}m$ )

0%
$7\ fm$
0%
$8\ fm$
0%
$9\ fm$
0%
$10\ fm$

Explanation

Given here,

$q_{1} = 120 \times e$

$q_{2} = e$ ( charge on proton)

It is based on conservation of energy of proton.

$\dfrac{1}{2}mv^2 = k \dfrac{q_1q_2}{R}$

$v = \sqrt{\dfrac{2Kq_1q_2}{mR}}$

Also, we know, de’broglie wavelength,

$\lambda =\dfrac{h}{mv}$

$\lambda = \dfrac{h\sqrt{R}}{\sqrt{2mKq_1q_2}}\\$

Thus,

$\lambda = \dfrac{h\sqrt{10\times10^{-15}}}{\sqrt{2\times\dfrac{5}{3}\times10^{-27}\times9\times10^9\times120\times e^2}}\\$

$\lambda =\dfrac{\dfrac{h}{e}\times10^{-7}}{6\times10^{-8}}\\$

$\lambda =\dfrac{4.2\times10^{-15}\times10^{-7}}{6\times10^{-8}}\\$

$\lambda =7\times10^{-15}\ m = 7\ fm \\$

Option A is correct.

The light of radiation

$300 nm$ falls on a photocell operating in the saturation mode. The spectral sensitivity is

$4.8 mA/W.$ The yield of photo electrons (i.e. number of electrons produced per photon) is

0%
$0.04$
0%
$0.02$
0%
$0.03$
0%
$0.2$

Explanation

energy per photon per second =

$\dfrac{hC}{\lambda}$

$\dfrac{6.663\times 10^{-34}\times 3\times 10^8}{3\times 10^{-11}}$

$6.6\times 10^{-19} J/s$
=

$6.6\times 10^{-19} W$

$1 W$ gives

$4.8 mA$
then

$6.6\times 10^{-15}$ W gives =

$(4.8\times 10^{-3}\times 6.6\times 10^{-19})A$

$= 31.7\times 10^{-21}A$
assuming n no. of electrons per second gives

$31.7\times 10^{-8} A$ of current, then

$I = \dfrac{\Delta Q}{\Delta t}$

$31.7\times 10^{-21} = \dfrac{n\times 1.6\times 10^{-19}}{1}$

$n = 0.2$

The de-Broglie wavelength of a neutron at

$927^{0}$ C is

$\lambda$ . Its wavelength at

$27^{0}$ C is:

0%
$\dfrac{\lambda}{2}$
0%
$\lambda$
0%
$2\lambda$
0%
$4\lambda$

Explanation

$\lambda = \dfrac{h}{\sqrt{2mKT}}$

$T_1 = 927 + 273$ ,

$T_2 = 27 + 273$

$= 1200 \ K$ ,

$= 300\ K$

$\lambda \alpha \dfrac{1}{\sqrt{T}}$

$\therefore \dfrac{{\lambda}_1}{{\lambda}_2} = \dfrac{\sqrt{T_2}}{\sqrt{T_1}}$

$\dfrac{\lambda_1}{{\lambda}_2} = \dfrac{\sqrt{300}}{\sqrt{1200}}$

$\ {{\lambda}_2 = 2 \lambda}$

Choose the correct statements from the following about photoelectric emission

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For given emitter illuminated by light of a given frequency, the number of photoelectrons emitted per second is proportional to the intensity of incident light
0%
For every emitter there is a definite threshold frequency below which no photo electrons are emitted, no matter what the intensity of light is
0%
Above the threshold frequency, the maximum kinetic energy of photo electrons is proportional to the frequency of incident light
0%
The saturation value of the photoelectric current is independent of the intensity of incident light

Explanation

D : Saturation current

$\alpha$ Integrity of incident light
C :

$KE = hV - \phi$ ;

$KE \alpha V$
B : True. if

$hV > \phi$ , then no photoelectron is emitted.
A : No. of photon

$\uparrow$ , no. of photoelectron

$\uparrow$
[provided frequency of photon is greater than threshold frequency]

When photons of energy

$4.25 eV$ strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy of

$K_{A}$ eV and de Broglie wavelength

$\lambda_{A}$ . The maximum kinetic energy of photo electrons liberated from another metal B by photons of energy

$4.70 eV$ is

$K_{B}=(K_{A}-1.5) eV$ . If the de Broglie wavelength of these photoelectrons is

$\lambda _{B}=2\lambda _{A}$ then

0%
the work function of A is 2.25 eV
0%
the work function of B is 4.20 eV
0%
$K_{A}=2.00$ eV
0%
$K_{B}=2.75$ eV

Explanation

De Broglie wavelength

$\lambda=\dfrac{h}{p}$ and kinetic energy,

$K=\dfrac{p^2}{2m}$

Thus,

$\dfrac{K_A}{K_B}=\dfrac{P_A^2}{P_B^2}=\dfrac{\lambda_B^2}{\lambda_A^2}=4$ as

$\lambda_B=2\lambda_A$

Given,

$K_B=(K_A-1.5)=4K_B-1.5 \Rightarrow K_B=0.5 eV$

Thus,

$K_A=4\times 0.5=2.00 eV$

From photo-electric effect:

Work function of A

$=4.25-K_A=4.25-2.00=2.25 eV$

and work function of B

$=4.70-K_B=4.70-0.5=4.20 eV$

Thus, only the option D is wrong.

Select correct alternative :
Statement -1 : An electric dipole can not produce zero electric field at a point which is situated at finite distance from the dipole.
Statement -2 : Mass of a moving photon is proportional to

$\lambda^{-1}$ .
Statement -3 : Equation of continuity for fluid is based on conservation of volume.

0%
FFF
0%
TTF
0%
TFF
0%
TTT

Explanation

Statemen-1: The potential due to a electric dipole of moment (p) is:

$\phi=\dfrac {1}{4\pi \epsilon} \dfrac{\vec{p}.\hat{r}}{r^2}$
The electric field from a dipole can be found from the gradient of this potential which cannot be zero at a point which is situated at finite distance from the dipole.

Statemen-2: From de-Brogli formula: The mass of a moving electron is given by:

$m=\dfrac{h}{\lambda v}$ . So, mass of a moving photon is proportional to

$\lambda^{-1}$

Statemen-3: A continuity equation in physics is an equation that describes the transport of a conserved quantity. Since mass, energy, momentum, electric charge and other natural quantities are conserved under their respective appropriate conditions, a variety of physical phenomena may be described using continuity equations.

So the answer is TTF.

The number of photons emitted per second by a

$40$ W lamp, if

$15 \%$ of the energy appears as the radiation of wavelength 540

$\mathrm { nm }$ is

0%
$1.63 \times 10 ^ { 19 }$
0%
$1.5 \times 10 ^ { 10 }$
0%
$2.2 \times 10 ^ { 10 }$
0%
$1.63 \times 10 ^ { 18 }$

A photosensitive metallic surface has work function, h

$_{0}$ . If photons of energy 2h

$_{0}$ fall on this surface, the electrons come out with a maximum velocity of 4 10

$^{6}$ m/s. When the photon energy is increased to 5h

$_{0}$ , then maximum velocity of photo electrons will be

0%
2 10 $^{7}$ m/s
0%
2 10 $^{6}$ m/s
0%
8 10 $^{5}$ m/s
0%
8 10 $^{6}$ m/s

Explanation

$2hv_{0}=hv_{0}+\frac{1}{2}m(4\times 10^{6})^{2}$

$5hv_{0}=hv_{0}+\frac{1}{2}mv^{2}\Rightarrow 4 \times \frac{1}{2}m(4\times 10^{6})$

$=\frac{1}{2}mv^{2}\Rightarrow v = 8 \times 10^{6} m/s$

A point source of light of power

$P$ and wavelength

$\lambda$ is emitting light in all directions. The number of photons present in a spherical region of radius

$r$ to radius

$r+x$ with centre at the source is:

0%
$\dfrac{P\lambda}{4 \pi r^2 hc}$
0%
$\dfrac{P\lambda x}{hc^2}$
0%
$\dfrac{P\lambda x}{4 \pi r^2 hc}$
0%
None of these

Explanation

The energy stored in the spherical region between

$r$ and

$r+x$ is

= energy density

$\times 4 \pi r^2 x=\dfrac{I}{C} \times 4 \pi r^2 x=\dfrac{P}{4 \pi r^2 c}4 \pi r^2 x=\dfrac{Px}{c}$

$\therefore$ Number of photons is the given volume is

$=\dfrac{Px}{h \nu c}=\dfrac{P\lambda x}{hc^2}$

A charge particle

$q_0$ of mass

$m_0$ is projected along the y-axis at

$t = 0$ from origin with a velocity

$V_0.$ If a uniform electric field

$E_0$ also exists along the x-axis, then the time at which de Broglie wavelength of the particle becomes half of the initial value is :

0%
$\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$
0%
$2\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$
0%
$\sqrt3\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$
0%
$3\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$

Explanation

Initial debrogle wavelength of the charge particle

$\lambda_0=\dfrac{h}{m_0v_0}$
After any time t,

$\lambda'=\dfrac{h}{m_{0}\sqrt{V_{0}^{2}+\left ( \dfrac{qE_{0}t}{m_{0}} \right )^{2}}}$
Suppose after time t,

$\lambda$ becomes half of the initial value i.e.

$\lambda=\lambda_0 /2$ .

$t=\sqrt3 \dfrac{m_0v_0}{q_0E_0}$

A particle of mass

$'m\ '$ is projected from ground with velocity

$'u\ '$ making an angle

$'\theta \ '$ with the vertical. The de-Broglie wavelength of the particle at the highest point is

0%
$\infty$
0%
$\dfrac{h}{mu sin \theta}$
0%
$\dfrac{h}{mu cos \theta}$
0%
$\dfrac{h}{mu}$

Explanation

The particle of depicted in adjacent figure.
Velocity at highest point

$= u sin \theta$
So, de-Broglie wave length at that point

$\lambda_D=\dfrac{h}{mu sin \theta}$ .

If the shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is

$0.01nm$ , then the de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately

$(hc = 12400 e VA, h = 6.63 \times 10^{-34}$ in MKS, mass of electron

$= 9.1 \times 10^{-31}kg)$

0%
$0.35$
0%
$0.035$
0%
$35$
0%
$1350$

Explanation

The shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is

$0.01nm$ .

So,

$\lambda_{min} = \dfrac{12400}{V}\Rightarrow V=\dfrac{1240}{0.01}=124000 volts$

The de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately

$\Rightarrow \lambda = \sqrt{\dfrac{150}{V}} = 0.035$

A small 50 kg vehicle is designed to be moved in free space by a lamp which emits 100 watts of red light of

$\lambda = 6630 A^o$ . What is its acceleration?

0%
$6.66 \times 10^{-9}\ m/sec^{2}$
0%
$3 \times 10^{-9}\ m/sec^{2}$
0%
$2.22 \times 10^{-9}\ m/sec^{2}$
0%
$4.44\ m/sec^{2}$

Explanation

In this problem it is the particle nature of light which determines the behavior of space vehicle. Since the emitted light carries away momentum and since the latter is carried, the rocket will be propelled to the left as shown in Figure.

Let N photons be emitted per sec.Each photon carries energy hv.
Then, power of lamp

$p=\dfrac{Energy\hspace{0.1in}of\hspace{0.1in}the\hspace{0.1in}photon \times number\hspace{0.1in}of\hspace{0.1in}photons}{time}=$

$Nhv$
But

$v=\dfrac{c}{\lambda }$

$\therefore p=\dfrac{Nhc}{\lambda }$ or

$N=\dfrac{\rho \lambda }{hc}=\dfrac{100\times 6630\times 10^{-10}}{6.63\times 10^{-34}\times 3\times 10^{8}}$

$=\dfrac{663\times 10^{-7}}{663\times 10^{-28}\times 3}=\frac{1}{3}\times 10^{21}=3.33\times 10^{20}$
Each particle will have momentum by de Broglie's hypothesis

$\displaystyle \mathrm{p}=\frac{\mathrm{h}}{\lambda}$

$\dfrac{6.63\times 10^{-34}}{6630\times 10^{-10}}=10^{-27}$ N-sec
Total force

$=\dfrac{d(n\phi )}{dt}$ or

$\displaystyle \mathrm{F}=\mathrm{p}\frac{\mathrm{d}\mathrm{n}}{\mathrm{d}\mathrm{t}}=$ p.N

$\mathrm{F}=10^{-27}\times 3.33\times 10^{20}=3.33\times 10^{-7}$ newton
Acceleration

$\dfrac{Force}{mass}=\dfrac{3.33\times 10^{-7}}{50}=6.66\times10^{-9}m/sec^{2}$

A photon of frequency v has an energy

0%
$\frac {h}{v}$
0%
$\frac {v}{h}$
0%
v
0%
hv

If a hydrogen atom at rest, emits a photon of wavelength

$\lambda ,$ the recoil speed of the atom of mass

$'m'$ is given by

0%
$\dfrac{h}{m\lambda }$
0%
$\dfrac{mh }{\lambda }$
0%
$mh\lambda$
0%
none of these

Explanation

From conservation of momentum,

$P = const = \vec{p}_{photon}+\vec{p}_{atom}=0$
Now, from de-broglie:

$\lambda =\dfrac{h}{mv}\Rightarrow v=\dfrac{h}{m\lambda }$

So, the answer is option (A).

Photoelectric effect is described as the ejection of electrons from the surface of metal when

0%
It is heated to high temperature
0%
Light of suitable wavelength falls on it
0%
Electrons of a suitable velocity impinge on it
0%
It is placed in a strong magnetic field

Photoelectric effect can be explained only by assuming that

0%
Light is a form of transverse waves
0%
Light is a form of longitudinal waves
0%
Light can be polarised
0%
Light consists of quanta

A light of wavelength

$\lambda$ is incident on a metal sheet of work function

$\phi = 2eV$ . The wavelength

$\lambda$ varies with time as

$\lambda = 3000 + 40t, where \lambda$ is in and t is in second. The power incident on metal sheet is constant at 100 W. This signal is switched on and off for time intervals of 2 minutes and 1 minute respectively. Each time the signal is switched on, the

$\lambda$ start from an initial value of 3000. The metal plate is grounded and electron clouding is negligible. The efficiency of photoemission is 1%

$(hc = 12400 eV)$ . The time after which photo-emission will stop is

0%
79 s
0%
80 s
0%
81 s
0%
78 s

Explanation

We have,

$\lambda = \dfrac{hc}{\phi}$

$\therefore \lambda=6200 A^{o}$
Also,

$\lambda= 3000+40t$

$\therefore t=80s$

So, the answer is option (B).

A proton and an electron are accelerated by the same potential difference. Let

$\lambda _{e}$ and

$\lambda _{p}$ denote the de Broglie wavelengths of the electron and the proton respectively.

0%
$\lambda _{e}=\lambda _{p}$
0%
$\lambda _{e}< \lambda _{p}$
0%
$\lambda _{e}> \lambda _{p}$
0%
$\lambda _{e}$ and $\lambda _{p}$ depend on the accelerating potential difference.

Explanation

Charge present on both panticle are same

$\because \lambda = \dfrac{h}{mqV}$ so

$q_1 = q_2$

$\dfrac{\lambda _{e}}{\lambda _{p}}=\sqrt{\dfrac{m_{p}v}{m_{e}v}}=>1$

$\because m_{p}> >m_{e}$

$\therefore \lambda _{e}> \lambda _{p}$ .

If the shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is

$0.1\mathring{A}$ , then the de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately
(

$hc=12400eV\mathring{A}$ ,

$h=6.63\times{10}^{-34}$ in MKS, mass of electron

$=9.1\times{10}^{-31}kg$ )

0%
$0.35\mathring{A}$
0%
$0.035\mathring{A}$
0%
$35\mathring{A}$
0%
$3.5\mathring{A}$

If we assume that penetrating power of any radiation/particle is inversely proportional to the De-broglie wavelength of the particle, then

0%
A proton and an $\alpha-$ particle after getting accelerated through same potential difference will have equal penetrating power.
0%
Penetrating power of $\alpha-$ particle will be greater than that of proton which have been accelerated by same potential difference.
0%
Proton's penetrating power will be less than penetrating power of an electron which has been accelerated by the same potential difference.
0%
Penetrating powers can not be compared as all these are particles having no wavelength or wave nature.

Explanation

$\lambda = \dfrac{h}{P} = \dfrac{h}{\sqrt{2mE}} \Rightarrow \dfrac{h}{\sqrt{2m(Vq)}}$

$\therefore$ For higher m and q ;

$\lambda$ will be smaller.
For an '

$\alpha$ ' particle; both 'm' and 'q' are higher
hence lesser is the wavelength.
As, (penetrating power)

$\propto$ Energy

$\propto \dfrac{1}{\lambda}$
From above; penetrating power of an

$\alpha-$ particle is more than that of a proton.

So, the answer is option (B).

A photon of frequency v has a momentum associated with it. If c is the velocity of light, then momentum is

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$\frac {hv}{c^2}$
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$\frac {hv}{c}$
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$\frac {v}{c}$
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$h vc$

A photon is an

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Quantum of light energy
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Quantum of matter
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Positively charged particle
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instrument for measuring light intensity

Two particles of masses

$m$ and

$2m$ have equal kinetic energies. Their de Broglie wavelengths are in the ratio of :

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$1 : 1$
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$1 : 2$
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$1:\sqrt{2}$
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$\sqrt{2}:1$

Explanation

de Broglie wavelength is

$\lambda = \dfrac{h}{p}=\dfrac{h}{\sqrt{2mk}}$ where

$k=$ kinetic energy

$\Rightarrow \dfrac{\lambda _{1}}{\lambda _{2}}= \dfrac{\sqrt{m_{2}k_{2}}}{\sqrt{m_{1}k_{1}}}=\sqrt{\dfrac{m_{2}}{m_{1}}}= \dfrac{\sqrt{2}}{1}$

$\dfrac{\lambda _{1}}{\lambda _{2}}=\dfrac{\sqrt{2}}{1}$

Will the radiation from a 50 kW, 100 MHz FM station expose the film?

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No
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Yes
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Cannot Say
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Incomplete data

Explanation

$f= 100 MHz = 10^8 Hz$

$\lambda = \dfrac{c}{f}= \dfrac{3 \times 10^8}{10^8} =3 m$

$E = hf = \dfrac{hc}{\lambda} = 413 \times 10^{-9} eV =660.8 \times 10^{-27} J$

Hence, cannot dissociate

$AgBr$ .

So, the answer is option (A).

The idea of quantum nature of light has emerged in an attempt to explain

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The thermal radiation of a black body
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The interference of light
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Radioactivity
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Thermionic emission

Find the frequency of photon.

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2.71 $\times$ 10 $^{14}Hz$
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2.01 $\times$ 10 $^{14}Hz$
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2.5 $\times$ 10 $^{14}Hz$
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20.1 $\times$ 10 $^{14}Hz$

Explanation

The energy of the incident photon that is just able to dissociate an AgBr molecule is,

$E =\dfrac{10^{5}}{6.023\times 10^{23}\times 1.6\times 10^{19}}=1.04eV$

The wavelength of the photon is,

$\displaystyle \lambda =\dfrac{1240}{1.04}=1200nm$

Frequency,

$\displaystyle \nu=\dfrac{C}{\lambda }=\frac{3\times10^{8}}{1200\times10^{-9}}=2.5\times10^{14}Hz$ .

So, the answer is option (C).

Mass of the photon at rest is

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$1.67\times 10^{-35} kg$
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One a.m.u
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$9\times 10^{-31} kg$
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Zero

de Broglie relation is true for

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All particles
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Charged particles only
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Negatively charged particles only
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Massless particles like photons only

Explanation

For any particle of mass m moving with velocity v has a de Broglie wavelength given by:

$\lambda=h/p$ . This is true for all material particle.

So, the answer is option (A).

A ray of energy 14.2 Me V is emitted from a

$^{60}Co$ nucleus. The recoil energy of the co-nucleus is nearly

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$3 \times 10^{-10}J$
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$3 \times 10^{-15}J$
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$3 \times 10^{-14}J$
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$3 \times 10^{-13}J$

Explanation

$P_r = P_{co'}$

$P_r = \dfrac{E}{c}$ ,

$KE = \displaystyle \dfrac{P^2}{2m} = \dfrac{E^2}{2c^2m_{co}}$

$=\displaystyle \dfrac{(14 \times 1.6 \times 10^{-13})^2}{2 \times(3\times 10^8 \times 60 \times 1.67 \times 10^{-27}}$

$= 3 \times 10^{-10}J$

So, the answer is option (A).

A proton and an electron are accelerated by the same potential difference. Let

$\displaystyle \lambda _{c}$ and

$\displaystyle \lambda _{p}$ denote the de Broglie wavelengths of the electron and the proton respectively.

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$\displaystyle \lambda _{c}= \lambda _{p}$
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$\displaystyle \lambda _{c}< \lambda _{p}$
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$\displaystyle \lambda _{c}> \lambda _{p}$
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The relation between $\displaystyle \lambda _{c}$ and $\displaystyle \lambda _{p}$ depends on the accelerating potential difference.

Explanation

de Broglie wavelength of a particle of mass m moving under a potential V is given my:

$\lambda=\dfrac{h}{\sqrt{2meV}}$ .

So it is clear that de Broglie wavelength is inversely proportional to the square root of the mass of the material. Now mass of electron is less that the mass of the proton so de Broglie wavelength of electron will be greater than that of proton.

So, the answer is option (C).

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 12 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 12 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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