JEE Main 2019 — Physics solution

JEE Main2019Physics

The stopping potential V₀ (in volt) as a function of frequency ( $ν$ ) 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^–19 C)

A
1.95 eV
B
2.12 eV
C
1.82 eV
D
1.66 eV

Solution

$ϕ = \frac{h c}{λ} = h v ∴ ϕ = h \times 4 \times 1 0^{14} H z = 1.654 e V \Rightarrow ϕ \approx 1.66 e V$

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