Physics

Electrons and Photons

127 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 wave optics huygens wave theory and wavefront wave propagation (huygens' construction) theories on light wave behaviour

Particle nature and wave nature of electomagnetic waves can be shown by

  1. electron has small mass, deflected by the metal sheet

  2. X-ray is diffracted a, reflected by thick metal sheet

  3. light is refracted and diffracted

  4. light is polarised and shows photoelectric effect.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Polarization is a property unique to transverse waves (wave nature), while the photoelectric effect demonstrates the particle nature of light.

Multiple choice physics electric current, potential difference and resistance electric potential and potential difference potential difference current in electric circuits

When you flip a switch to turn on a light, the delay before the light turns on is determined by :

  1. The speed of the electric fields moving in the wire.

  2. the drift speed of the electrons in the wire.

  3. the number of electron collisions per second in the wire.

  4. none of these, since the light comes instantly.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a switch is closed, the electric field propagates through the circuit at nearly the speed of light, causing electrons everywhere in the wire to start drifting almost simultaneously. The tiny delay before the bulb lights up is set by how fast the electromagnetic field travels through the conductors, not the very slow drift speed of the electrons themselves.

Multiple choice photosystems photosynthesis in plants metabolism, cell respiration, and photosynthesis photosynthesis in higher plants biology

After having absorbed the radiant energy by the pigment system I, electron is released by

  1. P ${ _6}$${ _8}$${ _3}$
  2. P ${ _6}$${ _7}$${ _3}$
  3. P ${ _7}$${ _0}$${ _0}$
  4. P ${ _6}$${ _8}$${ _0}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

P ${ _7}$${ _0}$${ _0}$ is the reaction center of the PS-I. So, when PS- I receives light energy electron is released by P ${ _7}$${ _0}$${ _0}$.

Multiple choice physics semiconductors band theory of solids, a brief introduction electron energies in solids energy bands

If the energy gap of a semiconductor is 1.1 e V it would be:

  1. Transparent to the ultraviolet radiation

  2. Opaque to the visible light

  3. Transparent to the visible light

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Visible light lies in the range of about 2.0 eV to 3.2 eV on the electromagnetic spectrum. 

This energy is sufficient to excite the valence electrons in the semiconductor and is hence absorbed by it. 
As a result, photons of lower energy are emitted which do not fall in the visible range of light. 
This causes the opacity of the semiconductor to visible light.

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

A light bulb has the resistance of tungsten resistance which convert about 10% electrical energy into visible light.  If energy other than visible is waste energy. How many kilo-joules does a light bulb wasted in one hour?

  1. 36 kJ

  2. 90 kJ

  3. 3240 kJ

  4. 360 kJ

  5. 32,400 kJ
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$90$ % of electrical energy is wasted in one hour.
Using        $1kWh  = 3600$ $kJ$
$\therefore$ Energy wasted       $E _{waste} = 0.9 \times 3600  =3240$  $kJ$
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

A copper ball of radius 1 cm work function 4.47 eV is irradiated with ultraviolet radiation of wavelength $2500\mathring { A } $. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be finite value of the potential on the ball. The charge acquired by the ball is :

  1. $5.5\times { 10 }^{ -13 }C$
  2. $7.5\times { 10 }^{ -13 }C$
  3. $4.5\times { 10 }^{ -12 }C$
  4. $2.5\times { 10 }^{ -11 }C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From photo electric effect equation :

]
$h\nu=h\nu _{0} + K.E _{max}$


so Maximum kinetic energy will be


$K.E _{max}= \dfrac{hc}{\lambda} - h\nu _{0}$
 
putting the given values in the above equation

$K.E _{max} = e\times V$ 

so V will be 

$V= \dfrac{k\times Q}{r}$
 
:: $ q = 5.5\times 10^{-13} C $

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

The photoelectric cut off voltage in a certain experiment is 1.5 V. The maximum kinetic energy of photoelectrons emitted is then

  1. 2.4 eV

  2. 1.5 eV

  3. 3.1 eV

  4. 4.5 eV
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The minimum negative potential applied to the plate or anode  for which the photoelectric current just becomes zero, So, in this case, maximum K.E of an electron will be equal to stopping potential.
Here, $V _0 \, = \, 1.5 \, V,$
Maximum Kinetic energy = $eV _{0} \, = \, 1.5 \,eV$

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

Light of wavelength 0.6 mm from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 mm from a sodium lamp, the stopping potential is 1.5 V. With this data, the value of h/e is:

  1. $6\times {{10}^{-5}}\,V{{s}^{-1}} $
  2. $2\times {{10}^{-15}}\,V{{s}^{-1}} $
  3. $4\times {{10}^{-55}}\,V{{s}^{-1}} $
  4. $4\times {{10}^{-15}}\,V{{s}^{-1}} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In emission of electron, Potential Energy

$eV = \dfrac{hc}{\lambda} - W _0$

When light of wavelength $\lambda =0.6\,mm$and stopping potential $0.5\,V$
$0.5e=\dfrac{hc}{6\times {{10}^{-7}}}-{{W} _{0}}\ ......\ (1)$

When light of wavelength $\lambda =0.4\,mm$and stopping potential $1.5\,V$
$1.5e=\dfrac{hc}{4\times {{10}^{-7}}}-{{W} _{0}}\ ......\ (2)$
subtract equation (1) from (2) $ e=\dfrac{hc}{{{10}^{-7}}}\left[ \dfrac{1}{4}-\dfrac{1}{6} \right] $

$ \Rightarrow \dfrac{h}{e}=\dfrac{12\times {{10}^{-7}}}{3\times {{10}^{8}}}=4\times {{10}^{-15}}\,V{{s}^{-1}} $ 

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

A photo-electric threshold wavelength of tungsten is $2300 \mathring{A}$ . The energy of electrons ejected from the surface, if ultra -violet light of wavelength $1800 \mathring{A}$ is incident on it, is 

  1. 1.5 eV

  2. 2 eV

  3. 3.2 eV

  4. 6 eV
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First calculate the work function energy corresponding to the threshold wavelength: phi = hc / lambda_0 = (12400 eV Angstrom) / 2300 Angstrom approx 5.39 eV (or use hc = 1240 eV nm). Then calculate the energy of the incident photon: E = hc / lambda = 12400 / 1800 approx 6.89 eV. The maximum kinetic energy of the ejected electrons is K_max = E - phi = 6.89 - 5.39 = 1.5 eV.

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

A metalic surface is irradiated with monochr matic light of variable wavelength. Above wavelength of $ 5000 \dot { A }  $ , no photoelectrons a emitted from the surface. With an unknown wavelength, a stopping potential of 3V is nessary to eliminate the photo current . The Known wavelength is:

  1. 2258 $ \dot { A } $
  2. $ 4133 \dot { A } $
  3. $ 3126 \dot { A } $
  4. $ 2679 \dot { A } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work function phi = 12400 / 5000 = 2.48 eV. Stopping potential 3V means max KE = 3 eV. Total energy E = 2.48 + 3 = 5.48 eV. Wavelength = 12400 / 5.48 = 2262 Angstroms. Option A is the closest.

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

The maximum energy of emitted photo electrons is measured by

  1. The current they produce

  2. The potential difference they produce

  3. The largest potential difference they can traverse

  4. The speed with which they emerge
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Maximum energy of emitted photo electrons is equal to the stopping potential. Stopping potential is the largest potential difference an emitted photo electron can traverse.

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

If the work function of the metal is W and the frequency of the incident light is v , then there is no emission of photo-electrons if

  1. $v< W/h$
  2. $v > W/h$
  3. $v \geq W/h$
  4. $v \leq W/h$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Maximum Kinetic Energy $=h\nu -\phi $
$=h\nu -W $
Since there is no emission
So, $h\nu -W < 0$
or $\nu < \dfrac{W}{h}$

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

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

  1. $3.2 \times 10^{-21}$
  2. $1.6 \times 10^{-17}$
  3. $3.2 \times 10^{-19}$
  4. $3.2 \times 10^{-15}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Energy\quad of\quad radiation,\quad h\nu =6.2eV\ =6.2\times 1.6\times { 10 }^{ -19 }J\ Work\quad function\quad W=4.2eV\ =4.2\times 1.6\times { 10 }^{ -19 }J\ KE=h\nu -W\ =6.2\times 1.6\times { 10 }^{ -19 }-4.2\times 1.6\times { 10 }^{ -19 }\ =2\times 1.6\times { 10 }^{ -19 }\ =3.2{ \times 10 }^{ -19 }J$