Question

A lightof 4 eV incident on metal surface of work function a; eV. Another light of 2.5 eV incident on another metal surface of work function «.: eV. Find the ratio of maximum velocity of photo electrons | v1/v2 I

Answer 1

A light of 4 eV incident on metal surface of work function Φ₁ eV. another light of 2.5 eV incident on another metal surface of work function Φ₂ eV.

To find : the ratio of maximum velocity of photo electrons (v1/v2)

solution : using photoelectric effect,

1/2 mv² = E - Φ

where Φ is work function, E is incident energy, m is mass and v is velocity of photo electron.

case 1 : E = 4 eV , Φ = Φ₁

so, 1/2 mv₁² = 4 - Φ₁.......(1)

case 2 : E = 2.5 eV , Φ = Φ₂

so, 1/2 mv₂² = 2.5 - Φ₂ .......(2)

dividing equations (1) and (2) we get,

⇒(v₁/v₂)² = (4 - Φ₁)/(2.5 - Φ₂)

⇒v₁/v₂ = √{(4 - Φ₁)/(2.5 - Φ₂)}

Therefore the correct option is (2)

Answer 2

Answer:

option 2

Explanation:

$\boxed{\bf Given,}$

⇒A light of 4 eV incident on metal surface of work function Ф₁ eV.

   Another light of 2.5 eV incident on another metal surface of work function Ф₂ eV.

$\bf {Formula:}\\\\ \boxed{ KE=h\nu -h\nu_0}\\\\ where \\ \it{KE = Kinetic \ energy \ of \ emitted \ electrons=\frac{1}{2}mv^2}\\ h\nu_0=Work \ function\\\\ \boxed{\frac{1}{2}mv^2=E-h\nu_0}\\\\\\ \rightarrow E_1=4 \ eV,h\nu_0_1=\phi _1eV\\\\ \frac{1}{2}mv_1^2= 4-\phi_1---> eq. 1\\\\ \rightarrow E_2=2.5 \ eV,h\nu_0_2=\phi_2 \ eV\\ \frac{1}{2}mv_2^2=2.5-\phi_2 --->eq.2\\\\ \text{dividing both equations}\\ \frac{v_1^2}{v_2^2}=\frac{4-\phi_1}{2.5-\phi_2}$

$(\frac{v_1}{v_2})^2 =\frac{4-\phi_1}{2.5-\phi_2} \\\\ \boxed{\frac{v_1}{v_2} = \sqrt{\frac{4-\phi_1}{2.5-\phi_2} }}$