When a certain photosensitive surface is illuminated with monochromatic light of frequency v , the stopping potential for the photo current is $-\mathrm{V}_{0} / 2$. When the surface is illuminated by monochromatic light of frequency $\mathrm{v} / 2$, the stopping potential is $-\mathrm{V}_{0}$. The threshold frequency for photoelectric emission is :

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Modern Physics

When a certain photosensitive surface is illuminated with monochromatic light of frequency v , the stopping potential for the photo current is

- V_{0} /2

. When the surface is illuminated by monochromatic light of frequency

v /2

, the stopping potential is

- V_{0}

. The threshold frequency for photoelectric emission is :

Answer: (D)

Use

e V_{s} = h ν - h ν_{t h}

with the magnitudes of the stopping potentials. For frequency

ν

e \frac{V _{0}}{2} = h ν - h ν_{t h}

; for frequency

\frac{ν}{2}

e V_{0} = h \frac{ν}{2} - h ν_{t h}

. Doubling the first,

e V_{0} = 2 h ν - 2 h ν_{t h}

, and equating the two expressions for

e V_{0}

gives

2 h ν - 2 h ν_{t h} = \frac{h ν}{2} - h ν_{t h}

. This collapses to

\frac{3}{2} h ν = h ν_{t h}

, so the threshold frequency is

ν_{t h} = \frac{3 ν}{2}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option