Question

Light of wavelength 3000 A falls on a metal surface having work function 2.3 ev. Calculate the
maximum velocity of ejected electrons.
(n = 6.63x10-34 J -s, C = 3x108 M/s and M=9.1x10-31Kg​

Answer 1

Answer:

The maximum velocity of ejected electrons is 8 × 10⁵ m/s

Explanation:

Given :

Light of wavelength 3000 Å falls on a metal surface having work function 2.3 eV

To find :

the maximum velocity of ejected electrons.

Solution :

λ = 3000 Å = 3000 × 10⁻¹⁰ m = 3 × 10⁻⁷ m

M = 9.1 x 10⁻³¹ kg

Φₒ = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ J

h = 6.63 x 10⁻³⁴ Js

c = 3 x 10⁸ m/s

We know,

KE = hc/λ – Φₒ

Substitute the values,

$KE = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^{8} }{3 \times 10^{-7}} - 2.3 \times 1.6 \times 10^{-19} \\\\ KE = 6.63 \times 10^{-34} \times 10^{8+7} - 3.68 \times 10^{-19} \\\\ KE = 6.63 \times 10^{-34+15} - 3.68 \times 10^{-19} \\\\ KE = 6.63 \times 10^{-19} - 3.68 \times 10^{-19} \\\\ KE = 2.95 \times 10^{-19} J$

KE = ½ × mv²

where v is the velocity

Substitute,

$2.95 \times 10^{-19} = \dfrac{1}{2} \times 9.1 \times 10^{-31} \times v^2 \\\\ v^2 = \dfrac{2.95 \times 10^{-19} \times 2}{9.1 \times 10^{-31}} \\\\ v^2 = \dfrac{5.9 \times 10^{-19+31}}{9.1} \\\\ v^2 = 0.64 \times 10^{12} \\\\ v= \sqrt{0.64 \times 10^{12}} \\\\ v = 0.8 \times 10^6 \\\\ \longrightarrow \tt v = 8 \times 10^5 m/s$

Therefore, the maximum velocity of ejected electrons is 8 × 10⁵ m/s