Question

Monochromatic light of wavelength λ = 4770 ˚A is incident separately on the surfaces of four different metals A, B, C and D. The work functions of A, B, C, and D are 4.2 eV, 3.7 eV,3.2 eV and 2.3 eV, respectively. The metal/ metals from which electrons will be emitted is/are:

• A. A, B, C and D
• B. A, B and C
• C. C and D
• D. D only

Hint:

For photoelectric emission, the photon energy must be greater than or equal to the work function of the metal

Answer 1

The Correct Option is D

1. Step 1: First, we need to calculate the energy of the photon using the equation:

   \( E = \frac{hc}{\lambda} \)

   Where \( h \) is Planck’s constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. Step 2: For the given wavelength of \( \lambda = 4770 \, \text{Å} \), we can calculate the energy of the photon as:

   \( E = \frac{(6.626 \times 10^{-34} \, \text{Js}) (3 \times 10^8 \, \text{m/s})}{4770 \times 10^{-10} \, \text{m}} = 4.15 \, \text{eV} \)

3. Step 3: Electrons will be emitted from a metal surface only if the energy of the incident photon is greater than or equal to the work function (\( \phi \)) of the metal.
4. Step 4: Comparing the photon energy with the work functions, we find that only metal D (with a work function of \( 2.3 \, \text{eV} \)) will emit electrons because its work function is less than or equal to the energy of the photon.

Conclusion: Metal D will emit electrons as its work function (\( 2.3 \, \text{eV} \)) is less than the photon energy (\( 4.15 \, \text{eV} \)).

Answer 2

Calculate the energy of the incident photons

Convert the wavelength to nanometers:

λ = 4770 Å = 477 nm

Calculate the photon energy:

E = hc/λ = (1240 eV⋅nm) / (477 nm) ≈ 2.6 eV

Ans : (D)

hc / λ = E = 2.6 eV > work function for photoelectrons emission