The electric field of light wave is given as $\overset{\to}{E} = {10}^{– 3}\cos \left(\dfrac{2\pix}{5 \times {10}^{– 7}} - 2\pi \times 6 \times {10}^{14}t\right)\overset{^}{x}\dfrac{N}{C}$ . This light falls on a metal plate of work function $2eV$ . The stopping potential of the photo-electrons is:
Given, $E$ (in $eV$ ) $= \dfrac{12375}{\lambda ( inÅ )}$

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The electric field of light wave is given as

\overset{\to}{E} = {10}^{– 3}\cos \left(\dfrac{2\pix}{5 \times {10}^{– 7}} - 2\pi \times 6 \times {10}^{14}t\right)\overset{^}{x}\dfrac{N}{C}

. This light falls on a metal plate of work function

2eV

. The stopping potential of the photo-electrons is:
Given,

E

(in

eV

)

= \dfrac{12375}{\lambda ( inÅ )}

Quantrex Academy · Accepted Answer

Correct answer: D. 0.48V

→ E = 10 ^ - 3 [ 2 5 × 10 ^ - 7 - 2π × 6 × 10 ^ 14 t] ^ x N / c k = 2π 5 × 10 ^ - 7 And ω = 6 × 10 ^ 14 × 2π ⇒ f = 6 × 10 ^ 14 Hz E = ϕ + K E _ K E _ = hf - ϕ = 6.6 × 10 ^ - 34 × 6 × 10 ^ 14 1.6 × 10 ^ - 19 - 2 = 2.475 - 2 = 0.48eV ∴ Stopping potential is 0.48V

The electric field of light wave is given as → E = 10 ^ – 3 ( 2 5 ×...

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