The electric field of light wave is given as $\vec{E} = 10^{- 3} \cos (\frac{2 π x}{5 \times 10^{- 7}} - 2 π \times 6 \times 10^{14} t) \hat{x} \frac{N}{C}$ . This light falls on a metal plate of work function $2 e V$ . The stopping potential of the photo-electrons is:
Given, $E$ (in $e V$ ) $= \frac{12375}{λ (i n Å)}$

Question

The electric field of light wave is given as

\vec{E} = 10^{- 3} \cos (\frac{2 π x}{5 \times 10^{- 7}} - 2 π \times 6 \times 10^{14} t) \hat{x} \frac{N}{C}

. This light falls on a metal plate of work function

2 e V

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

E

(in

e V

)

= \frac{12375}{λ (i n Å)}

Aicharya · Accepted Answer

E → = 10 - 3 c o s 2 π x 5 × 10 - 7 - 2 π × 6 × 10 14 t x ^ N / c k = 2 π 5 × 10 - 7 And ω = 6 × 10 14 × 2 π ⇒ f = 6 × 10 14 H z E = ϕ + K E m a x K E m a x = h f - ϕ = 6.6 × 10 - 34 × 6 × 10 14 1.6 × 10 - 19 - 2 = 2.475 - 2 = 0.48 e V ∴ Stopping potential is 0.48 V

The electric field of light wave is given as   E   → = 10 – 3 cos ⁡ 2 π x 5…

Solution

The electric field of light wave is given as &#160; E &#160; &#8594; = 10 &#8211; 3 cos &#8289; 2 &#960; x 5…

Solution

The electric field of light wave is given as E → = 10 – 3 cos ⁡ 2 π x 5…