Question

incident UV Light having wavelength lambda 1 is incident on metal surface and the ejected electron has the velocity of v1 .Another uv light having wavelength lambda 2 is incident on same metal surface ejects electron having vellocity v2 then v2squre -v1squre is

Answer 1

Answer:

(2hc/m)[( 1/lambda2 )- ( 1/ lambda1)]

Answer 2

Answer:

The value of (v2² - v1²) = (2hc/m)(1/λ2 -1/λ1 ) .

Explanation:

Given that:

• Wavelength of first UV light = λ1
• Wavelength of second UV light = λ2

Solution:

• The phenomenon of emission of electrons from metal surface when an electromagnetic radiation of high frequency is incident on it is called photoelectric effect .
• The photo generated electron are called photo electrons.
• Energy of the incident Photon = maximum kinetic energy of photoelectron + work function.

$E= \frac{1}{2 }m {v}^{2} +W$

• Let the work function of the metal surface be W.
• The energy of first incident UV light is E1 = hc/λ1
• Let, the maximum velocity of photoelectron be v1.
• Then, Energy of photoelectron is ½mv1².

$\frac{1}{2}m {v1}^{2} = \frac{hc}{λ1} - W \: \: - - - (1)$

• The energy of first incident UV light is E1 = hc/λ2
• Let, the maximum velocity of photoelectron be v2.
• Then, Energy of photoelectron is ½mv2².

$\frac{1}{2}m {v2}^{2} = \frac{hc}{λ2} - W \: \: - - - (2)$

• Subtracting (1) from (2), we get;

$\frac{1}{2}m {v2}^{2} - \frac{1}{2}m {v2}^{2} = \frac{hc}{λ2} - \frac{hc}{λ1} \\ {v2}^{2} - { v1}^{2} = \frac{2hc}{m} ( \frac{1}{λ2} - \frac{1}{λ1} )$

• Hence, the (v2² - v1²) = (2hc/m)(1/λ2 -1/λ1 ) .