# The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

Vikas TU
14149 Points
3 years ago
Given -
λ = 3300 A° = 330 nm
# Solution -
Energy released by given frequency is calculated by -
E = hc / λ
E = 1242 / 330 eV
E = 3.76 eV
Photoelectric emission occurs when incident energy is more than that of work function of metal.
As E is more than Φ in Na (2.75 eV) and K (2.30 eV), these metals will emit photons.
As E is less than Φ in Mo (4.17 eV) and Ni (5.15 eV), these metals will not show photoemission.
# If laser is brought nearer at 50 cm -
When laser is brought near at 50 cm, intensity of photoelectric emission will increase for Na and K.
Energy of incident radiation is unchanged with decrease in distance. Thus even after nearing laser, Mo and Ni will not show photoemission.