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Question:

The stopping potential V0 (in volt) as a function of frequency (v) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be :(Given : Planck's constant (h) = 6.63×10⁻³⁴ Js, electron charge e = 1.6×10⁻¹⁹C)

1.66 eV

2.12 eV

1.95 eV

1.82 eV

Solution:

Correct option is C. 1.66 eV
hv = φ + eV0
V0 = (hv/e) - (φ/e)
V0 is zero for v = 4×10¹⁴ Hz
0 = (hv/e) - (φ/e) => φ = hv
φ = (6.63×10⁻³⁴ × 4×10¹⁴) / (1.6×10⁻¹⁹) = 1.66 eV