The stopping potential V0 (in volt) as a function of frequency (n) 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)

Question

The stopping potential V0 (in volt) as a function of frequency (n) 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) 1.82 eV (C) 1.66 eV (D) 2.12 eV

Tardigrade Admin · Accepted Answer

hv = φ +ev0 v0 =(hv/e) - (φ/e) v0 is zero for v=4 ×1014 Hz 0 = (hv/e) - (φ/e) ⇒ φ = hv = (6.63 ×10-34 ×4 ×1014/1.6 ×10-19) = 1.66 ev

Q. The stopping potential V0​ (in volt) as a function of frequency (n) 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)

1.95 eV

1.82 eV

1.66 eV

2.12 eV

hv=ϕ+ev0​ v0​=ehv​−eϕ​ v0​ is zero for v=4×1014Hz 0=ehv​−eϕ​⇒ϕ=hv =1.6×10−196.63×10−34×4×1014​=1.66ev

$h v = ϕ + e v_{0}$
$v_{0} = \frac{h v}{e} - \frac{ϕ}{e}$
$v_{0}$ is zero for $v = 4 \times 1 0^{14} Hz$
$0 = \frac{h v}{e} - \frac{ϕ}{e} \Rightarrow ϕ = h v$
$= \frac{6.63 \times 1 0 ^{- 34} \times 4 \times 1 0 ^{14}}{1.6 \times 1 0 ^{- 19}} = 1.66 e v$