When a metallic surface is illuminated with r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Einstein's photoelectric equation,the stopping potential $V$ is given by $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lam

Vedclass · Accepted Answer

$3 \lambda$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the stopping potential $V$ is given by $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$For the second case: $e(\frac{V}{4}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(2)$Dividing equation $(1)$ by equation $(2)$:$4 = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}}$$4 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$$4(\frac{1}{2\lambda} - \frac{1}{\lambda_0}) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$\frac{2}{\lambda} - \frac{4}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$\frac{2}{\lambda} - \frac{1}{\lambda} = \frac{4}{\lambda_0} - \frac{1}{\lambda_0}$$\frac{1}{\lambda} = \frac{3}{\lambda_0}$$\lambda_0 = 3\lambda$.

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$,the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is:

Similar Questions

$Assertion (A)$: van der Waals' forces are responsible for chemisorption. $Reason (R)$: High temperature is favourable for chemisorption. The correct answer is

The total number of $RNA$ strands in an animal virus is:

If $A, B, C, D$ are angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D$ is equal to

The degree of ionization of $0.10 \ M$ lactic acid is $4.0 \ \%$. The value of $K_a$ is

The condition for the lines $lx + my + n = 0$ and $l_1x + m_1y + n_1 = 0$ to be conjugate with respect to the circle $x^2 + y^2 = r^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module