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 electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

The mean free path of electrons in a metal is $44 \times 10 ^ { - 8 } \mathrm { m }$ . Theelectric field which can give on an average 2$e \mathrm { V }$ energy to an electron in the metal will be in units of VIm 

  1. $8 \times 10 ^ { 7 }$
  2. $5 \times 10 ^ { - 11 }$
  3. $8 \times 10 ^ { - 11 }$
  4. $5 \times 10 ^ { 7 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

The drift velocity of electrons in a conducting wire is of the order of $1:mm/s$, yet the bulb glows very quickly after the switch is put on because

  1. the random speed of electrons is very high of the order of $10^{-6}m/s$

  2. the electrons transfer their energy very quickly through collision.

  3. electric field is set up in the wire very quickly, producing a current through each cross section, almost intantaneously

  4. All the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Drift velocity $v _d= \frac{eEt}{m}$ so drift velocity is directly proportional  to the electric filed. When switch is on, the filed is quick set up wire  and produce current through wire instantaneously.   

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
C Correct answer
Multiple choice luminous intensity measurements physics

A battery operated torch is adjusted to give a parallel beam of light. It produces illuminance of 60 lux on a wall 2m away. The illuminance produced 3 m away is 

  1. $60$ lux
  2. $\displaystyle\ \frac{80}{3}$ lux
  3. $40$ lux

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since $\dfrac {dF}{dA}$ is a constant, where $F$ is the illuminance and $A$ is the area of cross section. Hence the illuminance produced 3 m away does not change and it is 60 lux.

Multiple choice luminous intensity measurements physics

A battery-operated torch is adjusted to send an almost parallel beam of light. It produce an illuminance of $40 \ lux$ when light falls on a wall $2 m$ away. The illuminance produced when it falls on a wall $4 m$ away is close to

  1. $40\ lux$
  2. $20\ lux$
  3. $10\ lux$
  4. $5\ lux$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Illuminance =\dfrac{ Luminous \ Power} { Area}$


Since the beam of light is parallel, so the area does not change with distance. Also the power of torch being same, the luminous power remains same.

Therefore illuminance remains same on a wall 2 m away and 4 m away.

Illuminance produced on a wall 4 m away = 40 lux

Answer. A) 40 lux

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 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.