MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 9

Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?

Explanation

Hint:- According to de-Broglie hypothesis a wave is associated with a moving particle whose wavelength is given by

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

Explanation:-

As wavelength is given by

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

$\lambda \propto \dfrac{1}{P}$

$P \propto \dfrac {1}{\lambda}$

As momentum is inversely proportional to wavelength of light so graph between momentum and wavelength is hyperbolic.

The de Broglie wavelength

$\lambda$ associated with a proton increases by

$25\%$ . If its momentum is decreased by

$p_0$ . The initial momentum was:

0%
$4 p_0$
0%
$\dfrac{p_0}{4}$
0%
$5p_0$
0%
$\dfrac{p_0}{5}$

Explanation

Initial wavelength =

$\lambda$
final =

$\lambda +\dfrac{\lambda}{4}=\dfrac{5 \lambda }{4}$
Initial momentum =

$\dfrac{h}{\lambda}$
final momentum =

$\dfrac{4\lambda}{5 \lambda}$

$\Delta P=\dfrac{h}{5 \lambda}=P_0$
initial momentum =

$\dfrac{h}{\lambda}=5P_0$

A monochromatic beam of electromagnetic radiation has an intensity of

$1W/{m^2}$ . Then the average number of photons per

${m^3}$ for a 10MeV

$\gamma$ ray is ?

0%
4166
0%
3000
0%
5000
0%
2083

Explanation

We know,

$I=\dfrac{E}{t\times A}$

$I=\dfrac{nh\nu}{t\times A}$

given,

$h\nu=10\times 10^6\times 1.6\times 10^{-19}=1.6\times 10^{-12}$

Let

$l$ be distance travelled in time

$t$ ,

On multiplication with

$l$ in numerator and denominator

$1=\dfrac{nl1.6\times 10^{-12}}{tlA}=\dfrac{1.6\times 10^{-12}\times c}{lA}$

$\dfrac{n}{Al}=\dfrac{10^4}{4.8}=2083$

Option

$\textbf D$ is the correct answer

When light of intensity

$1\ W/m^{2}$ and wavelength

$5\times 10^{-7}m$ is incident on a surface, it is completely absorbed by the surface. If

$100$ photos emit one electron and area of the surface is

$1\ cm^{2}$ , then the photoelectric current will be

0%
$2\ mA$
0%
$0.4\ \mu A$
0%
$4.0\ mA$
0%
$4\ \mu A$

The De-Broglie wavelength associated with electrons revolving round the nucleus in a hydrogen atom in ground state, will be-

0%
$0.3\ A^{\circ}$
0%
$3.3\ A^{\circ}$
0%
$6.62\ A^{\circ}$
0%
$10\ A^{\circ}$

Explanation

According to

$Bohr's$ postulate

$mvr=\dfrac{nh}{2\pi}$ so the momentum in ground state

$(n=1)$ will be

$p=mv=\dfrac{h}{2\pi r}$

Required wavelength will be

$\lambda= \dfrac{h}{p}=\dfrac{h}{h/2 \pi r}=2\pi r$

Value of radius for ground state in hydrogen is

$r=0.53Angstrom$

Using that we get

$\lambda=3.32Angstrom$

Option B is correct.

The de Broglie wavelength corresponding to the root-mean-square velocity of hydrogen molecules at temperature

$20^{o}\ C$ is

0%
$3.3A^{o}$
0%
$1.04A^{o}$
0%
$126A^{o}$
0%
$34A^{o}$

The ratio of

$d-$ Broglie wavelength of molecular of hydrogen and oxygen kept in two vessels separately at

$27^oC$ and

$127^oC$ respectively is:

0%
$\sqrt{\dfrac{4}{3}}$
0%
$\sqrt{\dfrac{3}{32}}$
0%
$\dfrac{8}{\sqrt{3}}$
0%
$\dfrac{1}{8}$

Explanation

We know that,

$\dfrac { { \lambda H }_{ 2 } }{ { \lambda H }_{ e } } =\dfrac { \sqrt { { M }_{ { H }_{ e } }{ T }_{ { H }_{ e } } } }{ \sqrt { { M }_{ { H }_{ 2 } }{ T }_{ { H }_{ 2 } } } }$

Now,

$\dfrac { { \lambda H }_{ 2 } }{ { \lambda H }_{ e } } =\sqrt { \dfrac { 4\left( 127+273 \right) }{ 2\left( 27+273 \right) } }$

$\dfrac { { \lambda H }_{ 2 } }{ { \lambda H }_{ e } } =\sqrt { \dfrac { 1600 }{ 600 } }$

$\dfrac { { \lambda H }_{ 2 } }{ { \lambda H }_{ e } } =\sqrt { \dfrac { 8 }{ 3 } } =\sqrt { \dfrac { 4 }{ 3 } }$

When a point source of light is at a distance of

$50cm$ from a photoelectric cell, the cut -off voltage is found to be

$V_0$ .If the same source is placed at a distance of

$1m$ from the cell,then the cut -off voltage will be:

0%
$V_0/4$
0%
$V_0/2$
0%
$V_0$
0%
$2V_0$

Explanation

Shopping potential is Independent of distance (only Intensity depends upon distance). Hence ,New stopping potential mill remain unchanged

$\therefore$ It is

$V_0$

A proton is accelerated through

$225\, V$ . Its de Brogile wavelength is:

0%
$1.91\, pm$
0%
$0.2\, pm$
0%
$3\, pm$
0%
$0.4 \,nm$

Explanation

De-Brogile wavelength

$(\lambda)$ of a changed particle when accelerate through a potential

$v$ is given by:

$\lambda=\dfrac{h}{\sqrt{2mqv}}$ [

$h:$ Plank's constant]

in case of proton,

$m$ (man of proton)=

$1.67\times 10^{-27}Kg$

$q$ (charge of proton)=

$1.6\times 10^{-19}C$

Thus it gives:-

$[Given,V-225v]$

$\lambda=\dfrac{6.62\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.6\times 10^{-19}\times 225}}$

$=\dfrac{6.62\times 10^{-34}}{\sqrt{1.2\times 10^{-43}}}$

$\Rightarrow \lambda \simeq 1.91\times 10^{-12}m=1.91pm.$

what is the angular momentum of an electron of de

$-$ broglie wavelength

$\lambda ?$ given

$r$ is the radius of orbit

$.$

0%
$\dfrac{{rh}}{\lambda }$
0%
$\dfrac{{2rh}}{\lambda }$
0%
$\dfrac{{3rh}}{\lambda }$
0%
$\dfrac{{4rh}}{\lambda }$

Explanation

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

$r = \dfrac{{rh}}{\lambda }$

Hence,

option

$(A)$ is correct answer.

According to de Broglie, wavelength of electron in second orbit is

$10^{-9}$ meter. Then the circumference of orbit is :-

0%
$10^{-9}$
0%
$2 \times 10^{-9}$
0%
$3 \times 10^{-9}$
0%
$4 \times 10^{-9}$

Threshold wavelength of tungsten is 2300 angstrom. If ultraviolet light of wavelength 1800 angstrom is incident on it, then the maximum kinetic energy of photoelectrons would be ?

0%
1.5 eV
0%
2.5eV
0%
3.0 eV
0%
5.0eV

Explanation

$k_{max}=hv-hv_o$

$=hc\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda}\right)$

$=1.24\times 10^{-6}\left(\dfrac{10^8}{18}-\dfrac{10^8}{23}\right)$

$k_{max}=1.489eV$

$k_{max}\approx 1.5ev$

Hence option A is correct.

A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to

0%
$H$
0%
$H\dfrac{1}{2}$
0%
$H^o$
0%
$H\dfrac{-1}{2}$

Explanation

Solution :

Velocity

$v= \sqrt{2gH}$

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

$= \dfrac{h}{m\sqrt{2gh}}$

$\lambda \alpha H^{-1/2}$

1207733_1136896_ans_3f27c7413b704a8ebf3941000441c433.png

A photon and an electron both have wavelength

$1A^{o}$ . The ratio of energy of photon to that of electron is

0%
$1$
0%
$0.012$
0%
$82.7$
0%
$10^{-10}$

The de-Broglie wavelength of an electron is the same as the wavelength of a photon. The K.E of photon is

$'x'$ times the K.E of the electron,then

$'x'$ is (m-mass electron ,h- planck's constant,c-velocity of light)

0%
$\dfrac{hc}{2\lambda m}$
0%
$\dfrac{2m\lambda c}{h}$
0%
$\dfrac{2\lambda c}{hm}$
0%
$\dfrac{2\lambda m}{ch}$

Explanation

$\dfrac{hc}{\lambda}=x\left(\dfrac{1}{2}mv^2\right)$

$\dfrac{hc}{\lambda}=x\left(\dfrac{1}{2}m\left(\dfrac{h}{m\lambda}\right)^2\right)^2$

$=\dfrac{hc}{\lambda}=x\left(\dfrac{nh^2}{2m^2\lambda 2}\right)$

$=x=\dfrac{hc}{\lambda}\times \dfrac{2m \lambda ^2}{h^2}$

$=x=\dfrac{2m\lambda c}{h}$

The radio of deBroglie wavelengths of a proton and an alpha particle of same energy is

0%
$1$
0%
$2$
0%
$4$
0%
$0.25$

A chlorine molecule with an initial velocity of

$600ms^{-1}$ absorbs a photon of wavelength 350 nm and is then dissociated into two chlorine atoms. One of the atoms is detected moving perpendicular to the initial direction of the molecule and having a velocity of

$1600ms^{-1}$ . The binding energy of the molecule is (approx)? [Neglect the momentum of the absorbed photon. The relative atomic mass of chlorine is 35]

0%
$3.36\times 10^{-17}J$
0%
$3.36\times 10^{-19}J$
0%
$3.36\times 10^{-21}J$
0%
$3.36\times 10^{-28}J$

Electron has energy of 100

${ e }{V }$ what will be its wavelength

0%
$1.2{ \mathring { A } }$
0%
$10 { \mathring { A } }$
0%
$100 { \mathring { A } }$
0%
$1 \mathring { A }$

The de Broglie wavelength of an electron moving with a velocity

$1.5 \times{ 10 }^{ 8 }{ ms }^{ -1 }$ is equal to that a photon. The ratio of the kinetic energy of the electron to the that of the photon is

0%
2
0%
4
0%
$\dfrac { 1 }{ 2 }$
0%
$\dfrac { 1 }{ 4 }$

A photon and an electron both have wavelength

$1\mathring { A }$ . The ratio of energy of photon to that of electron is

0%
1
0%
0.012
0%
82.7
0%
${ 10 }^{ -10 }$

If the rest mass of an electron or positron is

$0.51\ MeV$ , then the kinetic energy of each particle, in the electron-positron pair produced by a

$\gamma-$ photon of

$2.42\ MeV$ , will be ?

0%
$0.3\ MeV$
0%
$1.9\ MeV$
0%
$0.7\ MeV$
0%
$1.5\ MeV$

Explanation

Initial energy is that of the

$\gamma$ photon which is

$E=2.42MeV$

Energy consumed in forming the electron-positron pair will be

$0.51MeV+ 0.51MeV=1.02MeV$

So remaining energy will be

$E^{ \prime}=2.42MeV-1.02MeV=1.4MeV$

$\text{This remaining energy will be equally divided in both particles as both have same mass i.e 1.4/2=0.7MeV }$

$\text { for each particle in electron-positron pair.}$

Option c is correct.

$\alpha$ particle is moving in a circular path of radius in the presence of magnetic field B. The de-Broglie wavelength associated with the particle will be

$(q\rightarrow charge\quad on\quad \alpha \quad particle)$

0%
$\cfrac { hr }{ qB }$
0%
$\cfrac { hB }{ qr }$
0%
$\cfrac { h }{ qBr }$
0%
$\cfrac { h{ r }^{ 2 } }{ qB }$

Explanation

We know,

Radius

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

and,

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

$\Rightarrow \ \lambda = \dfrac{h}{qBr}$

$\therefore$ De-Broglie wavelength

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

Hence, option "C" is correct.

A photon and an electron have equal energy

$E$ .

${\lambda _{photon}} \times {\lambda _{electron}}$ is proportional to:

0%
$E^{\dfrac{-3}{2}}$
0%
$\frac{1}{{\sqrt E }}$
0%
$\frac{1}{E}$
0%
Does not depend upon E.

Explanation

For Electron,

$E=\dfrac{mv^{2}}{2}\Rightarrow v=\left(\dfrac{2E}{me}\right)^{\dfrac{1}{2}}$

$\lambda$ electron

$=\dfrac{h}{m_{e}v}=\dfrac{h}{me\left(\dfrac{2E}{me}\right)^{\dfrac{1}{2}}}=\dfrac{h}{(2E me)^{\dfrac{1}{2}}}$

For Electron

$E=hv\Rightarrow r=\dfrac{E}{h}$

$\lambda$ photon

$=\dfrac{c}{r}=\dfrac{hc}{E}$

$\lambda$ photon

$+\lambda$ electron

$=\dfrac{hc}{E}\dfrac{h}{2^{\dfrac{1}{2}}(Eme)^{\dfrac{1}{2}}}\alpha E^{\dfrac{-3}{2}}$

De-Broglie wavelength associated with thermal neutrons at room temperature of

$27^{0}$ C is :

0%
$1.452\times 10^{-10}$ m
0%
$2.57\times 10^{-10}$
0%
$0.452\times 10^{-10}$ m
0%
$3.452\times 10^{-10}$ m

Explanation

Given,

$T=27+273=300K$

${ \text { Boltzmann's constant. } }$

${ k = 1.38 \times 10 ^ { - 23 } J m o l ^ { - 1 } k ^ { - 1 } }$

Kinetic energy of neutron at absolute temperature

$\text{ T is given by, }E=\dfrac{3}{2}kT$

${ \lambda = \dfrac { h } { \sqrt { 2 m E } } }$ $=\dfrac{h}{\sqrt{3mkT}}$

${ \lambda = \dfrac { 6.63 \times 10 ^ { - 34 } } { \sqrt { 3 \times 1.675 \times 10 ^ { - 27 } \times 1.38 \times 10 ^ { - 23 } \times 300 } } }$ $=1.45\times {{10}^{-10}}m\text{ }$

is $1.45\times {{10}^{-10}}m\text{ }$

A particle of mass M at rest decays into particles of masses

$m_{1} \ and \ m_{2}$ having non -zero velocities. The ratio of the de broglie wavelengths of the particles

$\lambda _{1}/\lambda _{2}$ is :

0%
$m _{1}/m _{2}$
0%
$m _{2}/m _{1}$
0%
1
0%
m1.m2

Explanation

When mass

$M$ at rest (initial momentum

$=0$ ) decays to smaller masses

${m}_{1}$ and

${m}_{2}$ , energies have same momentum (From conservation of momentum)

${ \vec { P } }_{ 1 }+{ \vec { P } }_{ 2 }=0\Rightarrow { P }_{ 1 }={ P }_{ 2 }$ (magnitude)

And de-Broglie wavelength

${ \lambda }_{ 1 }=\cfrac { h }{ { P }_{ 1 } } ;{ \lambda }_{ 2 }=\cfrac { h }{ { P }_{ 2 } }$ (H-Plancks constant)

Thus we get

${ \lambda }_{ 1 }={ \lambda }_{ 2 }\Rightarrow \cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =1$

The threshold frequency for a photosenstive metal is

$7 \times 10 ^ { 13 } \mathrm { Hz}$ . If light of frequency

$10 ^ {14 } \mathrm { Hz}$ is incident on this metal, then maximum kinetic energy of emitted electron is?

0%
$4 \times 10 ^ { -20 } \mathrm {J}$
0%
$8 \times 10 ^ { -20 } \mathrm {J}$
0%
$2 \times 10 ^ { -20 } \mathrm {J}$
0%
$10 ^ { -20 } \mathrm {J}$

Explanation

We know,

$h(\nu-\nu_0)KE_{max}$

Given,

$\nu_0=7\times 10^13$

$\nu=10^14$

$KE_{max}=6.632\times 3\times 10^{-34+13}=19.896\times 10^{-21}=1.9896\times 10^{-20}$

$KE_{max}=2\times 10^{-20}$

Option

$\textbf C$ is the correct answer

The ratio of wavelength of deutron and proton accelerated through the same potential difference will be-

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

A source S

$_{1}$ is producing 10

$^{15}$ photons per second of wavelength

$5000$

$A^o$ . Another source S

$_{2}$ is producing

$1.02 \times 10^{15}$ photons per second of wavelength

$5100$

$A^o$ .Then,

$\dfrac{(power \ of \ S_{2})}{(power \ of \ S_{1})}$ is equal to

0%
$0.98$
0%
$1.00$
0%
$1.02$
0%
$1.04$

If Bohr radius is

$r_{0},$ the corresponding de Broglie wavelength of the electron is

0%
$\left ( \dfrac{2\pi }{r_{o}} \right )$
0%
$\left ( \dfrac{r_{o} }{2\pi } \right )$
0%
$\left ( \dfrac{1}{2\pi r_{o}} \right )$
0%
${2\pi r_{o}}$

Explanation

Given that,

Radius $r={{r}_{0}}$

We know that,

$2\pi r=n\lambda$

Suppose $n=1$

Then,

$\lambda =2\pi {{r}_{0}}$

Hence, this is the required solution

Energy of a photon having wave number

$1.00 cm^{-1}$ is

0%
$6.62\times 10^{-34}$ J
0%
$1.99\times 10^{-23}$ J
0%
$6.62\times 10^{-32}$ J
0%
$6.62\times 10^{-36}$ J

Explanation

As energy of a photon,

$E=ht={ h }_{ c }$

Here,

$h=6.626\times { 10 }^{ -34\lambda }Js$

As wavelength

$\left( \lambda \right) =1\ { cm }^{ -1 }=0.01/m$

Speed of light,

$c=3\times { 10 }^{ 8 }m/s$

So, Energy

$=\dfrac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 0.01 }$

Energy of photon

$=1987.8\times { 10 }^{ -26 }J$

Energy of photon

$=1.99\times { 10 }^{ -23 }J$

Hence, this is the answer.

Five elements

$A, B, C, D and E$ have work functions

$1.2 eV, 2.4 eV, 3.6 eV , 4.8eV and 6.8 eV$ respectively. If light of wavelength

$4000 A ^ { \circ }$ is allowed to fall on these elements, then photoelectrons are emitted by

0%
$A, B and C$
0%
$A, B, C, D and E$
0%
$A and B$
0%
$Only E$

Explanation

The threshold wavelength,

$\lambda _0 =\dfrac{hc}{W}=\dfrac{12400}{W(ev)}Angstrom$

so for

$A$

$\lambda _0 =\dfrac{hc}{W}=\dfrac{12400}{1.2}Angstrom=10333A^o$

and for

$B$

$\lambda _0 =\dfrac{hc}{W}=\dfrac{12400}{2.4}Angstrom=5166.66A^o$

for

$C$

$\lambda _0 =\dfrac{hc}{W}=\dfrac{12400}{3.6}Angstrom=3444.44A^o$

for

$D$

$\lambda _0 =\dfrac{hc}{W}=\dfrac{12400}{4.8}Angstrom=2583.33A^o$

for

$E$

$\lambda _0 =\dfrac{hc}{W}=\dfrac{12400}{6.8}Angstrom=1823.52A^o$

We see that

$4000A$ is only less than

$\lambda_0$ for

$A$ and

$B$

An electron beam has a kinetic energy equal to

$100\ eV$ . Find its wavelength associated with a beam, if mass of electron is

$9.1 \times 10 ^ { - 31 }\ kg$ and

$1 \text { eV } = 1.6 \times 10 ^ { - 19 } \mathrm { J }$ . (Planck's constant is

$6.6 \times 10 ^ { - 34 } \mathrm { J-s } )$

0%
$24.6 \mathrm { A }^{o}$
0%
$0.12 \mathrm { A }^{o}$
0%
$1.2 A^{o}$
0%
$6.3 \mathrm { A }^{o}$

Photoeelectric work-function of a metal is

$1eV.$ Light of wavelength

$\lambda = 3000 \mathring { A }$ falls on it. The photoelectrons come out with maximum velocity:

0%
$10 m/s$
0%
$10^3 m/s$
0%
$10^4 m/s$
0%
$10^6 m/s$

Explanation

Solution

By Binstein's photo electric

$eq^{n}.$

$hr = hr_{0}+KE_{max}$

$\Rightarrow \frac{hc}{h} = hr^{0}+\frac{1}{2}mV^{2}$

Here,

$\Rightarrow \frac{6.6\times 10^{-31}\times 3\times 10^{8}}{3\times 10^{-7}}=1.0\times 10^{-19}J+\frac{9.k10^{-31}}{z}\times V^{2}$

$\Rightarrow 6.6\times 10^{-19}-1.6\times 10^{-19} = \frac{9.1\times 10^{-31}}{2}V^{2}$

$\Rightarrow 2\times \frac{5\times 10^{-19}}{9.1\times 10^{-31}} = V^{2}$

$\Rightarrow V = \sqrt{\frac{10^{13}}{9.1}}$

$\Rightarrow V \approx 10^{6}m/s$

option -D is correct.

1156223_1191290_ans_4eef1c11b8e8443aa36db08e8dbb6e6c.jpg

$K_1$ and

$K_2$ are the maximum kinetic energies of the photo electrons emitted when light of wave lengths

$\lambda_1$ and

$\lambda_2$ respectively are incident on a metallic surface. If

$\lambda_1 = 3\lambda_2$ then

0%
$K_1 > \dfrac{K_3}{3}$
0%
$K_1 < \dfrac{K_2}{3}$
0%
$K-1 > 3K_2$
0%
$K_2 = 3K_1$

When an electron de-excited back from

$(n+1)^th$ state to

$n^th$ state in a hydrogen like atoms, wavelength of radiation emitted is

$\lambda_{1}(n>>1)$ . In the same atom de-broglies wavelength associated with an electron in

$n^th$ state is

$\lambda_{2}$ . Then

$\dfrac{\lambda_{1}}{\lambda_{2}}$ is proportional to

0%
$\dfrac{1}{n}$
0%
$n$
0%
$n^{2}$
0%
$n^{3}$

When light is incident on surface, photo electrons are emitted. For photoelectrons:

0%
The value of kinetic energy is same for all
0%
Maximum kinetic energy do not depend on the
wave length of incident light
0%
The value of kinetic energy is equal to or less
than a maximum kinetic energy
0%
None of the above

Explanation

Kinetic energy of photo electron is

KE = hf - W = $\frac{hc}{\lambda}$ - W

h = plancks constant

f = frequency

$\lambda$ = wavelength of incident light

W = work function
The photo electrons are not emitted instantly. They collide several times with other electrons within the metal before coming out. Hence the value of kinetic energy is not the same. From above equation, Kinetic energy depends on wavelength of incident light.Also the value of kinetic energy is less than or equal to $hf$
$Hence, Option (B) is correct$

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium Which are at temperature

$27^oC$ and

$127^oC$ respectively is

0%
$\dfrac { 1 }{ 2 }$
0%
$\sqrt { \dfrac { 3 }{ 8 } }$
0%
$\sqrt { \dfrac { 8 }{ 3 } }$
0%
$1$

Explanation

As we know,

de-Broglie wavelength

$\lambda =\dfrac{h}{mv_{rms}}, rms$ velocity of a gas particle at the given temperature

$(T)$ is given as

$\dfrac 12 mv_{rms}^2=\dfrac 32 kT\Rightarrow v_{rms}=\sqrt{\dfrac{3\ kT}{m}}\Rightarrow mv_{rms}=\sqrt{3mkT}$

$\therefore \lambda =\dfrac{h}{mv_{rms}}=\dfrac{h}{\sqrt{3\ mkT}}$

$\Rightarrow \dfrac{\lambda_H}{\lambda_{He}}=\sqrt{\dfrac {m_{He} T_{He}}{m_H T_H}}=\sqrt{\dfrac{4(273+127)}{2(273+27)}}=\sqrt {\dfrac 83}$

The velocity of photons is proportional to (where

$v =$ frequency)

0%
$\dfrac { 1 } { \sqrt { v } }$
0%
$v ^ { 2 }$
0%
$\mathrm { v } ^ { 0 }$
0%
$\sqrt { v }$

$E_{1},E_{2},E_{3}$ and

$E_{4}$ the respectively kinetic energies of electron, decuteron proton and neutron having same de-Broglic wavelength, identity the correct in which those value would increase,

0%
$E_{2},E_{4},E_{3},E_{1}$
0%
$E_{1},E_{3},E_{4},E_{2}$
0%
$E_{2},E_{4},E_{1},E_{3}$
0%
$E_{3},E_{1},E_{2},E_{4}$

Find the maximum

$KE$ of photo electrons liberated from the surface of lithium by electron magnetic radiation whose electric component varies with time as

$\varepsilon =a\left( 1+\cos { \omega t } \right) \cos { { \omega }_{ 0 } } t$ where

$'a'$ is a constant

$\left(\dfrac {\omega}{2\pi}\right) =6\times 10^{14}s^{-1}$ and

$\left(\dfrac {\omega _{0}}{2\pi}\right) =0.66\times 10^{14}s^{-1}$ , work function of lithium

$=2.39\ eV$

0%
$0.72\ eV$
0%
$0.36\ eV$
0%
$0.603\ eV$
0%
$0.99\ eV$

A hydrogen atom emits a photon corresponding to an electron transition from n =5 to n=The recoil speed of hydrogen atom is almost (mass of proton =

$1.6\times 10^{-27}\, kg$ )

0%
$10\ ms^{-1}$
0%
$2\times10^{-2}\ ms^{-1}$
0%
$4\ ms{-1}$
0%
$8\times10^2\ ms^{-1}$

Explanation

${ P }_{ H-atom }={ P }_{ photon }=\dfrac { { E }_{ radiated } }{ c }$

$=\dfrac { 13.6\left[ \dfrac { 1 }{ { n }_{ 1 }^{ 2 } } -\dfrac { 1 }{ { n }_{ 2 }^{ 2 } } \right] eV }{ C }$

$1.6\times { 10 }^{ -27 }\times v=\dfrac { 13.6\left[ \dfrac { 1 }{ 12 } -\dfrac { 1 }{ { 5 }^{ 2 } } \right] \times 1.6\times { 10 }^{ -14 } }{ 3\times { 10 }^{ 6 } }$

$\boxed { v\approx 4m{ s }^{ -1 } }$

If the debroglie wave-length of

$He$ at

$927^{o}C$ is

$\lambda$ then de-brolie wave-length of

$CH_{4}(g)$ at

$27^{o}C$ will be

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

The wavelength

$\lambda_{e}$ of an electron and

$\lambda_{p}$ of a photon of same energy

$E$ are related by

0%
$\lambda_{p} \propto \lambda_{e}^{2}$
0%
$\lambda_{p} \propto \lambda_{e}$
0%
$\lambda_{p} \propto \sqrt {\lambda_{e}}$
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$\lambda_{p} \propto \dfrac {1}{\sqrt {\lambda_{e}}}$

Explanation

${ \lambda }_{ p }\alpha { \lambda }_{ e }^{ 2 }$

Wavelength of electron

${ \lambda }_{ e }=\dfrac { h }{ \sqrt { 2mE } }$ and proton

${ \lambda }_{ p }=\dfrac { { h }_{ c } }{ E }$

$\Rightarrow { \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2mE }$ or

$E=\dfrac { { h }_{ c } }{ { \lambda }_{ p } }$

$\therefore$

${ \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2m\dfrac { { h }_{ c } }{ { \lambda }_{ p } } }$

$\Rightarrow { \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2m{ h }_{ c } } { \lambda }_{ p }$

${ \lambda }_{ e }^{ 2 }\alpha { \lambda }_{ p }$

An electron is moving in

${ 2 }^{ nd }$ excited orbit of H-atom Radius of orbit in terms of de-Broglie wavelength

$\lambda$ of electron can be given as

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$\cfrac { \lambda }{ \pi }$
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$\cfrac { 2\lambda }{ \pi }$
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$\cfrac { 3\lambda }{ 2\pi }$
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$\cfrac { \lambda }{ 2\pi }$

Explanation

From Bohr's postulate,

$mvr=\dfrac{nh}{2\pi}$ . . . . . .(1)

de-broglie wavelength,

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

$mv=\dfrac{h}{\lambda}$ . . . . . .(2)

Substitute equation (2) in equation (1), we get

$r\dfrac{h}{\lambda}=\dfrac{nh}{2\pi}$

$r=\dfrac{n\lambda}{2\pi}$

For 2nd excited state,

$n=3$

Radius of orbit,

$r=\dfrac{3\lambda}{2\pi}$

The correct option is C.

The work function of a metal is

$1.84\ eV$ . If the wavelength of incident photon is

$4000 \mathring{A}$ , Then the approximate maximum kinetic energy of emitted photoelectron will be

0%
$1.25\ eV$
0%
$3.1\ eV$
0%
$0.58\ eV$
0%
$1.8\ eV$

The debroglie wavelength of photoelectrons is

$1\mathring {A}$ . Its accelerating potential is :

0%
$150\ V$
0%
$15.3\ V$
0%
$12.3\ V$
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$13.6\ V$

A parallel beam of uniform, monochromatic light of wavelength 2640

$\overset { o }{ A }$ has an intensity of 100

$W/m^2$ . The number of photons in 1

$mm^3$ of this radiation are -

0%
266
0%
335
0%
442
0%
555

Explanation

Wavelength

$=2640A$

Intensity

$=$ Power / Area

$=200$

Area

$=1{ mm }^{ 2 }={ 10 }^{ -6 }{ m }^{ 2 }$

Energy of

${ 10 }^{ -6 }{ m }^{ 2 }$ area

$=200\times { 10 }^{ -6 }=2\times { 10 }^{ -4 }J/s$

No. of photons

$=2\times { 10 }^{ -4 }/7.5\times { 10 }^{ -19 }=2.66\times { 10 }^{ 14 }$

The momentum of a photon is

$2\times 10^{-16}$ gm-cm/sec. Its energy is

0%
$0.61\times 10^{-26}$ erg
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$2.0\times 10^{-26}$ erg
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$6\times 10^{-6}$ erg
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$6\times 10^{-8}$ erg

Explanation

Given,

$P=2\times 10^{-16} gm-cm/sec$

$c=3\times 10^{10}cm/sec$

Energy,

$E=Pc$

$E=2\times 10^{-16}\times 3\times 10^{10}$

$E=6\times 10^{-6} erg$

The correct option is C.

The de Broglie wavelength of a gas molecule at a temperature T K is:

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$\cfrac{h}{\sqrt{3mKT}}$
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$\cfrac{h}{3mKT}$
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$\cfrac{h}{\sqrt{2mKT}}$
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$\sqrt{2mKT}$

A particle of mass m kg and charge q coulomb is accelerated from rest through V volt: then the de-Broglie wavelength

$\lambda$ associated with it is given by

0%
$\lambda = \frac{h}{\sqrt{mV}}$
0%
$\lambda = \frac{h}{\sqrt{2mq}}$
0%
$\lambda = \frac{h}{\sqrt{2mqV}}$
0%
$\lambda = \frac{h}{\sqrt{2mV}}$

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 9 - MCQExams.com

Kinetic energy of photo electron is

KE = hf - W = $\frac{hc}{\lambda}$ - W

h = plancks constant

f = frequency

$\lambda$ = wavelength of incident light

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 9 - MCQExams.com

Kinetic energy of photo electron is

KE = hf - W = \frac{hc}{\lambda}\frac{hc}{\lambda} - W

h = plancks constant

f = frequency

\lambda\lambda= wavelength of incident light

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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KE = hf - W = $\frac{hc}{\lambda}$ - W

$\lambda$ = wavelength of incident light