Question

The work functions for metals M₁ and M₂ are 1.9 eV and 5.0 eV respectively. Perform necessary calculations to find out which metal emits photoelectrons, when monochromatic light of wavelength 410 nm is incident on them.

Answer 1

Given: Wavelength of incident light (λ) = 410 nm

= 410 × 10⁻⁹ m

Formula: E = `(hc)/λ`

= `(6.63 xx 10^-34 xx 3 xx 10^8)/(410 xx 10^-9)`

= `(19.89 xx 10^-26)/(410 xx 10^-9)`

= 0.0485 × 10⁻¹⁷

= 4.85 × 10⁻¹⁹ J

J convert to eV 

= `(4.85 xx 10^-19)/(1.6 xx 10^-19)`

= 3.03 eV

We compare the calculated photon energy (3.03 eV) with the given work functions:

For Metal M₁ (Φ₁) = 1.9 eV

Since 3.03 eV > 1.9 eV (Photon Energy > Work Function), the incident light has sufficient energy to eject electrons.

For Metal M₂ (Φ₂) = 5.0 eV

Since 3.03eV < 5.0eV (Photon Energy < Work Function), the incident light does not have enough energy to overcome the metal's work function.