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Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Ultraviolet radiation of 6.2eV falls on an aluminium surface (work function 4.2eV). The kinetic energy in joule of the faster electron emitted is approximately

  1. $3.2\times 10^{21}$

  2. $3.2\times 10^{-19}$

  3. $3.2\times 10^{-17}$

  4. $3.2\times 10^{-15}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$E _k=E-W _o=6.2-4.2=2.0 e V$
$=2.0 \times 1.6 \times 10^{-19}=3.2\times 10^{-19}\, J$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The ratio of de-Broglie wavelengths of proton and $\alpha$-particle having same kinetic energy is

  1. $\sqrt2 :1$

  2. $2\sqrt2 :1$

  3. 2 :1

  4. 4 : 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

De Broglie wavelength is given by:


$\lambda = \dfrac{h}{p}$ 

Writing momentum as a a function of kinetic energy and mass

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

i.e.   $\dfrac{\lambda _{proton}}{\lambda _{alpha}} = \sqrt{\dfrac{m _{alpha}}{m _{proton}}}$

$m _{alpha} = 4m _{proton}$
So,

$\dfrac{\lambda _{proton}}{\lambda _{alpha}} = \sqrt{\dfrac{m _{alpha}}{m _{proton}}} = \sqrt{\dfrac{4}{1}} = 2$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

$K _{1} $and $K _{2}$ are the maximum kinetic energies of the photoelectrons emitted when light of wavelength $\lambda _{1} $ and $\lambda _{2} $  respectively are incident on a metallic surface. If $\lambda _{1}= $3$\lambda _{2} $  then

  1. $K _{1}>\dfrac{K _{2}}{3}$

  2. $K _{1}<\dfrac{K _{2}}{3}$

  3. $K _{1}=3K _{2}$

  4. $K _{2}=3K _{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$K _{1}=\dfrac{hc}{\lambda _{1} }-\phi $-----------(1)

$K _{2}=\dfrac{hc}{\lambda _{2}}-\phi$-----------(2)

$\because\lambda _{1} = 3 \lambda _{2}$

$k _{2}=3\left(\dfrac{hc}{\lambda _{1}}\right)-\phi$---------(3)

$\dfrac{k _{2}}{3}=\dfrac{hc}{\lambda _{1}}-\dfrac{\phi}{3}$----------(4)

$\dfrac{k _{2}}{3}=(k _{1}+\phi)-\dfrac{\phi}{3}$  (by (1))

$\dfrac{k _{2}}{3}=k _{1}+\dfrac{2 \phi}{3}$

So,$k _{1}< \dfrac{k _{2}}{3}$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The work function of a metal is $1.6\times 10^{-19}$J. When the metal surface is illuminated by the light of wavelength 6400 $A^{o}$, then the maximum kinetic energy of emitted photoelectrons will be ($h = 6.4 \times 10^{-34} Js$)

  1. $14\times 10^{-19}J$

  2. $2.8\times 10^{-19}J$

  3. $1.4\times 10^{-19}J$

  4. $1.4\times 10^{-19}eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$K.E. _{max}=\dfrac{hc}{\lambda }-\phi $


$=\dfrac{6.4\times 10^{-34}\times 3\times 10^{8}}{6.4\times 10^{-7}}-1.6\times 10^{-19}$

$=3\times 10^{-19}-1.6\times 10^{-19}$
$=1.4\times 10^{-19}J.$
So, the answer is option (C).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The work function of a metal is 4.6eV. The wavelength of incident light required to emit photo-electrons of zero energy from its surface, will be

  1. 5000 $A^{0}$

  2. 3100 $A^{0}$

  3. 1700 $A^{0}$

  4. 2700 $A^{0}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$E=\dfrac{hc}{\lambda}$

$ 4.6eV=\dfrac{1240eV}{\lambda}$

        $\lambda=2700{A}^{0}$
Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The photoelectric work function of a metal surface is 2eV. When light of frequency $1.5 \times10^{15}$ Hz is incident on it, maximum kinetic energy of the photo-electrons, approximately is :

  1. 8 eV

  2. 6 eV

  3. 2 eV

  4. 4 eV
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$K.E. _{max}=h\nu-\phi $


=$\dfrac{6.6\times 10^{-34}\times 1.5\times 10^{15}}{1.6\times 10^{-19}}eV-2eV$

$=6eV-2eV$

$=4eV.$

So, the answer is option (D).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Work function of a metal is 3.0eV. It is illuminated by a light of wavelength $3 \times 10^{-7}$m. Then the maximum energy of the electron is.

  1. $2.34 \ eV$

  2. $0.85 \ eV$

  3. $1.13 \ eV$

  4. $3.32 \ eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Maximum \ \ energy =\dfrac{hc}{\lambda}-\phi  (3\times 10^{-7}m=300nm)$

$=\dfrac{1240}{300} - 3 \ \ \ \ (hc =  1240 \ eV / X nm)$
$=4.13-3$
$=1.13 eV.$
So, the answer is option (C).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The energy of the incident photon is 12.38 eV, while the energy of the scattered photon is 9.4 eV. The K.E. of the recoil electron is nearly

  1. 2 eV

  2. 1 eV

  3. 4 eV

  4. 3 eV
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

K.E of the recoil electron
= energy of the incident photon - energy of scattered photon
= 12.38 eV -9.4eV
$\simeq 3eV.$

So, the answer is option (D).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Light of wavelength 5000 $A^{o}$ falls on a sensitive plate with photoelectric work function 1.9eV. The maximum kinetic energy of the photoelectrons emitted will be

  1. $0.58 \ eV$

  2. $2.48 \ eV$

  3. $1.24 \ eV$

  4. $1.16 \ eV$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\lambda =5000 A^{0}  =  500nm$

$K.E. _{max} =\dfrac{hc}{\lambda }-\phi $

$=\dfrac{1240}{500}-1.9 \ \ \ \  (hc = 1240  eV - nm)$

$=2.48-1.9$
$=0.58 \ eV$
So, the answer is option (A).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

When light of wavelength 2480 $A^{0}$ is incident on a metal surface electrons are emitted with a maximum KE of 2 eV. The maximum KE of photo-electrons, if light of wavelength 1240 $A^{0}$ is incident on the same surface would be

  1. 4eV

  2. 1 eV

  3. 2eV

  4. 7eV
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From the 1st condition
$\dfrac{hC}{\lambda} - W = 2eV$

$h \ in \ terms \ of \ eV = \dfrac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}}$

$\Rightarrow \dfrac{4.14 \times 10^{-15} \times 3 \times 10^{8} \times 10^{10}}{2480}$

$\Rightarrow Work for = \left ( 5.012 - 2 \right )eV$

$\Rightarrow W \approx 3eV$

So,
In 2nd case when wavelength of incident light is $1240 A^{\circ}$,
$\Rightarrow \dfrac{hC}{\lambda} - W = K. E.$

$\Rightarrow K. E. = \dfrac{4.14 \times 10^{-15} \times 3 \times 10^{18}}{1240} - 3$

$=10 - 3 = 7eV$

So, the answer is option (D).