Question

If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?
Metal work function (eV)
Na. 1.92
K 2.15
Ca. 3.20
Mo. 4.17

Answer 1

Metal will show photoelectric emission only when energy of incident photons is greater than equal to work function of that metal.
[ Work function is the minimum amount of energy required to eject the electrons from metal surface. ]

Now, use formula , E = 1240/λ( in nm) eV
here , λ = 412.5 nm
So, E = 3.006 eV

Now, apply condition of photoelectric emission ,
E ≥ work function { Φ }
We can see that work function of K and Na are less than incident energy of photons . Hence, only potassium,K and Sodium ,Na will show photoelectric emission.

Answer 2

The correct answer is Na and K :

Explanation:

Photoelectric emission will be shown by metal if incident energy of photon is equal to or greater than the work function .
Energy of Incident photon E:

E=hν=hc/λ[joules]=hv/λ e [ eV]
=6.6.x10⁻³⁴ x 3x10⁸/412.5 x10⁻⁹ x 1.6 x 10⁻¹⁹E
=3ev

So according to given :

Metal work function (eV)
Na. 1.92
K 2.15
Ca. 3.20
Mo. 4.17
so  the incident photon energy is equal to or greater than work function only in case of sodium and Potassium.