# Photoelectric Effect

## Trending Questions

**Q.**What is the work function of the metal if a light of wavelength 4000 ∘A generates photoelectrons of velocity 6×105 ms−1 from it?

(Mass of electron = 9×10−31 kg, Velocity of light = 3×108 ms−1, Plank's constant = 6.626×10−34 Js, Charge of electron = 1.6×10−19 J eV−1)

- 3.1 eV
- 0.9 eV
- 4.0 eV
- 2.1 eV

**Q.**

When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68Ã—105J molâˆ’1. What is the maximum wavelength that will cause a photoelectron to be emitted?

- Î»=320 nm
- Î»=449.7 nm
- Î»=526 nm
- Î»=518 nm

**Q.**

Calculate energy of one mole of photons of radiation whose frequency as 5 × 1014Hz.

499.31 KJ

199.51 KJ

299.31 KJ

399.31 KJ

**Q.**A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

- 4.969×1020
- 2.012×1020
- 4.969×1018
- 2.012×1021

**Q.**

Calculate the velocity of the electron in the first Bohr orbit of the hydrogen atom (Given, r = a)

**Q.**

The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

**Q.**

A photon of wavelength 4 Ã— 10^{â€“7} m strikes on the metal surface, the work function of the metal is 2.13 eV. Calculate

(i) the energy of the photon (eV),

(ii) the kinetic energy of the emission, and

(iii) the velocity of the photoelectron (1 eV= 1.6020 Ã— 10^{â€“19} J).

**Q.**When the frequency of the incident radiation on a metallic plate is doubled, the kinetic energy of the photoelectrons will be:

- doubled
- halved
- more than doubled
- increases but less than doubled

**Q.**

How does the quantum theory explain the photoelectric effect?

**Q.**Ejection of a photoelectron from a metal in the photoelectric effect experiment can be stopped by applying 0.5 V when a radiation of 250 nm is used. The work function of the metal is :

- 4 eV
- 5.5 eV
- 4.5 eV
- 5 eV

**Q.**A certain metal when irradiated with light (ν=3.2×1016 Hz) emits photoelectrons with twice kinetic energy of photoelectrons that are emitted when the same metal is irradiated by light (ν=2.0×1016 Hz). Calculate νo of electron.

- 1.2 × 1014 Hz
- 8 × 1015 Hz
- 4 × 1012 Hz
- 1.2 × 1016 Hz

**Q.**Statement- 1: The kinetic energy of photoelectrons increases with increase in frequency of incident light.

Statement- 2: The number of photoelectrons ejected increases with increase in intensity of light.

- Both the statements are true.
- Statement- 1 is true and statement- 2 is false.
- Both the statements are false.
- Statement- 1 is false and statement- 2 is true.

**Q.**How does the intensity of the incident radiation affect photoelectric effect?

- As the intensity increases, photoelectric effect may increase
- As the intensity increases, photoelectric effect decreases.
- As the intensity decreases, photoelectric effect increases.
- No effect

**Q.**

An electron of mass ${\mathrm{m}}_{\mathrm{e}}$ and a proton of mass ${\mathrm{m}}_{\mathrm{p}}=$$1836{m}_{e}$ are moving at the same speed. The ratio of their de Broglie wavelength ${\mathrm{\xce\xbb}}_{\mathrm{electron}}/{\mathrm{\xce\xbb}}_{\mathrm{proton}}$ will be:

$918$

$1836$

$\frac{1}{1836}$

$1$

**Q.**The threshold frequency of a metal is 6×1014 s−1. Calculate the kinetic energy of an electron emitted when radiation of frequency 1.1×1015 s−1 hits the metal.

- 3.31 × 10−19 J
- 1.13 × 10−18 J
- 7.12 × 10−19 J
- None of These

**Q.**The energy of an electron in the second and the third Bohr's orbits of the hydrogen atom is −5.42×10−12 erg and −2.41×10−12 erg respectively. Calculate the wavelength of the emitted radiation when the electron drops from the third orbit to the second orbit.

- 6.604×103∘A
- 6.604∘A
- 3.905×103∘A
- 3.905∘A

**Q.**

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency ( v0 ) and work function ( W0 ) of the metal.

**Q.**The work function of metal is 4 eV. To emit a photo electron of zero velocity from the surface of metal, the wavelength of incident light should be:

- 5900 oA
- 2700 oA
- 1700 oA
- 3100 oA

**Q.**When radiation of λ=253.7 nm strikes the copper plate, the energy required to stop the ejection of electrons from copper plate is 0.24 eV. Calculate the work function of copper.

- 4.64 eV
- 6.65 eV
- 8.45 eV
- 10.25 eV

**Q.**

A Light of wavelength λ shines on a metal surface with intensity X, and the metal emits Y electrons per second of average energy Z. What will happen to Y and Z If X is doubled?

Y will be doubled and Z will become half.

Y will remain same and Z will be doubled.

Both Y and Z will be doubled.

Y will be doubled but Z will remain same.

**Q.**When a certain metal was irradiated with light of frequency 3.2×1016 Hz, the photoelectrons emitted had twice the kinetic energy as that of the photoelectrons emitted when the same metal was irradiated with light of frequency 2.0×1016 Hz. The value of ν∘ for the metal was found to be x×1016 Hz.

Then x=

**Q.**

When an electromagnetic radiation of wavelength 310 nm falls on the surface of sodium metal, electrons are emitted with K.Emax = 1.5 eV. Determine the work function Wo of metal.

- 2.5 eV
- 5.2 eV
- 10.5 eV
- 12.5 eV

**Q.**In a photoelectric experiment, the kinetic energy of photoelectrons was plotted against the frequency of the incident radiation (ν), as shown in the figure. Which of the following statements is correct?

- The slope of the line is equal to the Planck's constant.
- The threshold frequency is ν1
- As the frequency of the incident wavelength increases beyond the threshold frequency, the kinetic energy of the photoelectrons decreases.
- It is impossible to obtain such a graph.

**Q.**The maximum kinetic energy of photoelectrons ejected from a metal, when it is irradiated with radiation of frequency 2×1014Hz is 6.63×10−20J . The threshold frequency of the metal is: (Take the value of h= 6.63×10−34Jsec)

- 1×1014Hz
- 2×1014Hz
- 1×1015Hz
- 2×1015Hz

**Q.**In photoelectric effect, the kinetic energy of photoelectrons increases linearly with the

- Wavelength of incident light
- Frequency of incident light
- Velocity of incident light
- Atomic mass of an element.

**Q.**If the threshold wavelength (λ0) for ejection of electron from metal is 330 nm, then work function for the photoelectric emission is:

(Planck's constant (h) = 6.6×10−34 J s)

- 1.2×10−18 J
- 1.2×10−20 J
- 6×10−19 J
- 6×10−12 J

**Q.**The threshold frequency ν0 for a metal is 7.0×1014 s−1. Calculate the kinetic energy of an electron emitted when a radiation of frequency ν=1.0×1015 s−1 hits the metal.

- 1.988×10−19 J
- 4.567×10−19 J
- 1.988×10−16 J
- 4.567×10−16 J

**Q.**When the frequency of light incident on a metallic plate is doubled, the maximum kinetic energy of emitted photoelectrons will be:

- halved
- doubled
- unchanged
- increases but more than the double of previous kinetic energy

**Q.**When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68×105Jmol−1. What is the minimum energy needed to remove an electron from sodium?

- Emin=3.84×10−19J
- Emin=2.31×105J
- Emin=4.42×10−19J
- Emin=3.78×10−19J

**Q.**Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from the metal surface?