Question

Radiation of wavelength 331 nm irradiates the following metals:

Metal	Work Function (eV)
Na	1.92
K	2.15
Ca	3.20
Mo	4.17

Which of the following statements is correct?

पर्याय

• Only Na and K show photoelectric emission.

• Only Mo will not show photoelectric emission.

• All of the given metals show photoelectric emission.

• None of them show photoelectric emission.

Answer 1

Only Mo will not show photoelectric emission.

Explanation:

Only when the incident photon’s energy (E) is greater than or equal to the metal’s work function (Φ) does photoelectric emission take place.

Energy of a photon:

E = `(h c)/lambda`

In eV, using λ in nm:

E(eV) ≈ `1240/(lambda(nm))`

Incident wavelength (λ) = 331 nm

Energy of incident photons (E) = `1240/331`

≈ 3.746 eV

Comparing this energy with the given work functions: 

1. For Na (Φ = 1.92 eV): E > Φ, so emission occurs.
2. For K (Φ = 2.15 eV): E > Φ, so emission occurs.
3. For Ca (Φ = 3.20 eV): E > Φ, so emission occurs.
4. For Mo (Φ = 4.17 eV): E < Φ, so emission does NOT occur.

∴ The Na, K, and Ca show emission, while Mo does not. Thus, only Mo will not show photoelectric emission.