Physics

Electrons and Photons

158 Questions

Electrons and photons are fundamental concepts in modern physics, focusing on quantum interactions and the photoelectric effect. The topic involves calculating work functions, kinetic energy, and photon wavelengths. These physics questions are highly relevant for competitive test preparation.

Photoelectric effectWork function calculationsPhoton energy and wavelengthElectron emissionAtomic transitions

Electrons and Photons Questions

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

$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).

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

The cut-off voltage in a photoelectric experiment is 3V. Then the maximum KE of photo-electrons emitted is

  1. 3 V

  2. 3 eV

  3. 6 eV

  4. 9 eV
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Cut off voltage is the minimum voltage applied across the plates that even the electrons ejected with minimum kinetic energy could not reach the other plate.
So, 
From the definition,
Cut off voltage $= 3V$
Work done on the charge $=$ Kinetic energy of the photons
$\Rightarrow 3V \times 1e = K. E.$
$\Rightarrow K. E. = 3eV$

So, the answer is option (B).

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

If the frequency of light incident on a photosensitive metal plate is doubled, then the KE of photoelectrons will be

  1. Doubled

  2. Halved

  3. Quadrupled

  4. More than twice the previous value
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\ KE\ =\ h(V-V _{0})$
$\ V^{1}\ =\ 2V$
$\ KE^{1}\ =\ h(2V-V _{0})$
$\ KE^{1}\ =\ 2h(V-V _{0}) +\ hV _{0}$
$\ KE^{1}\ =\ 2KE+hV _{0}$

So, the answer is option (D).

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

The photo-electric threshold wavelength for a metal is $5000 A^{0}$. Light of wavelength $4000 A^{0}$ is incident on it. The maximum KE of photo-electrons emitted is [given $hc= 2 \times 10^{-25} Jm$]

  1. $3.1 eV$
  2. $2.48 eV$
  3. $0.62 eV$
  4. $5. 58 eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$K _{max}\ =\ \dfrac{hc}{\lambda}-\dfrac{hc}{\lambda _{0}}$

            $=\dfrac{20\times 10^{-26}}{4\times 10^{-7}}-\dfrac{20\times 10^{-26}}{5\times 10^{-7}}$

           $=\ 20\times 10^{-19}\left ( \dfrac{1}{4} -\dfrac{1}{5}\right )\ J$

          $=10^{-19}J=\dfrac{10^{-19}}{1.6\times 10^{-19}}eV=0.62eV$

So, the answer is option (C).

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

The work function of a certian metal is 2.3 eV .If light of wave number $2\times10^{6}m^{-1}$ falls on it,the kinetic energies of fastest and slowest ejected electorn will be respectively:

  1. 2.48eV ,0.18eV

  2. 0.18eV,Zero

  3. 2.30eV,Zero

  4. 0.18eV,0.18eV
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\phi _{ _0}=2.3eV$ 


$\lambda^{-1}=2\times10^6m^{-1}$


Some electrons, although get loosened due to the incident rays, remain stationary.
$\therefore KE _{min}=0$ always

$KE _{max}=TE-\phi _{ _0}$  where TE is the energy of the rays.

$TE=hc\lambda^{-1}$  in Joules

$TE=\dfrac{hc\lambda^{-1}}e$ $in$ $eV$ where e is the charge of one electron

$\therefore TE=\dfrac{6.63\times10^{-34}\times{3\times10^8}\times2\times 10^6}{1.6\times10^{-19}}=2.48eV$

$\therefore KE _{max}=2.48-2.3=0.18eV$

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

In photoelectric effect, initially when energy of electrons emitted is $E _{0}$, de-Broglie wavelength associated with them is $\lambda _{0}$. Now, energy is doubled then associated de-Broglie wavelength $\lambda^{'}$ is

  1. $\displaystyle\lambda^{'}=\frac{\lambda _{0}}{\sqrt{2}}$
  2. $\displaystyle\lambda^{'}=\sqrt{2}\lambda _{0}$
  3. $\displaystyle\lambda^{'}=\lambda _{0}$
  4. $\displaystyle\lambda^{'}=\frac{\lambda _{0}}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

de-Brogile wavelength is given by
$\displaystyle\lambda =\frac{h}{p}$, where h= Planck's constant and p= momentum
Also energy (E) and momentum are related as 
$\displaystyle E =\frac{p^{2}}{2m}$
$\displaystyle \therefore p=\sqrt{2mE}$
$\displaystyle \therefore \lambda =\frac{h}{\sqrt{2mE}}\times \frac{1}{\sqrt{E}}$ as h and m are constants
Hence, $\displaystyle \frac{\lambda _{0}}{{\lambda}'}=\sqrt{\frac{{E}'}{E}}=\sqrt{\frac{2E}{E}}=\sqrt{2}$
$\displaystyle \therefore  {\lambda}'= \frac{\lambda _{0}}{\sqrt{2}}$

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

In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is $1eV$ for incoming radiation of frequency $v _{0}$ and $3eV$ for incoming radiation of frequency $3v _{0}/2$. What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency $9v _{0}/4$?

  1. $3\ eV$
  2. $4.5\ eV$
  3. $6\ eV$
  4. $9\ eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$(KE) _{max} = hv - \phi _{0}$
So, $1\ eV = hv _{0} - \phi _{0} .... (i)$
and $3\ eV = \dfrac {hv _{0}}{2} - \phi _{0} .... (ii)$
$\Rightarrow 3\ eV - 1\ eV = \dfrac {hv _{0}}{2}$
or $hv _{0} = 4\ eV$
From Eq. (i), $\phi _{0} = hv _{0} - 1\ eV$
$= 4\ eV - 1\ eV = 3\ eV$
$\therefore (KE) _{mas} = h\times \dfrac {9v _{0}}{4} - 3\ eV$
$= \dfrac {9}{4} (4\ eV) - 3\ eV = 6\ eV$.