The stopping potential V0 (in volt) as a function of frequency (v) for a sodium emitter, is shown in the figure The work function (in eV) 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 (v) for a sodium emitter, is shown in the figure The work function (in eV) 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) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5897313a2f013bb13185c2a5b4625dae-.png" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f0=4× 1014 Hz W0 = hf0 = 6.63 × 10-34 × (4 × 1014) J (6.63 × 10-34 × 6 × 1014 /1.6 × 10-19) =1.66 eV

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

f0​=4×1014Hz W0​=hf0​=6.63×10−34×(4×1014)J 1.6×10−196.63×10−34×6×1014​ =1.66eV

$f_{0} = 4 \times 1 0^{14} Hz$
$W_{0} = h f_{0} = 6.63 \times 1 0^{- 34} \times (4 \times 1 0^{14}) J$
$\frac{6.63 \times 1 0 ^{- 34} \times 6 \times 1 0 ^{14}}{1.6 \times 1 0 ^{- 19}}$
$= 1.66 e V$