The electric field of light wave is given as $\vec{E}=10^{-3} \cos \left(\frac{2 \pi x}{5 \times 10^{-7}}-2 \pi \times 6 \times 10^{14} t\right) \hat{x} \frac{N}{C}$. This light falls on a metal plate of work function 2 eV . The stopping potential of the photoelectrons is:

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Modern Physics

The electric field of light wave is given as

E = 1 0^{- 3} cos (\frac{2 π x}{5 \times 1 0 ^{- 7}} - 2 π \times 6 \times 1 0^{14} t) \overset{x}{^} \frac{N}{C}

. This light falls on a metal plate of work function 2 eV . The stopping potential of the photoelectrons is:

(A)0.48 V✓ Correct
(B)2.48 V
(C)0.72 V
(D)2.0 V

Answer: (A)

Solution

The angular frequency in the wave is

ω = 2 π \times 6 \times 1 0^{14}

, so the frequency is

6 \times 1 0^{14}

Hz and the wavelength is

5 \times 1 0^{- 7}

= 5000

Angstrom. The photon energy is

E = \frac{h c}{λ} = \frac{12375 eV A ˚}{5000 A ˚} = 2.475

eV. By Einstein's equation the maximum kinetic energy is

2.475 - 2.0 = 0.475

eV, so the stopping potential is

0.475

\approx 0.48

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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