When radiation of wavelength lambda is incide | vedclass

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

Question

(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi_0$,where $\phi_0 = \frac{hc}{\lambda_0}$ and $\lambda_0$ is th

Vedclass · Accepted Answer

$4 \lambda$

Vedclass · Answer

(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi_0$,where $\phi_0 = \frac{hc}{\lambda_0}$ and $\lambda_0$ is the threshold wavelength.For the first case: $e(4.8) = \frac{hc}{\lambda} - \phi_0 \quad ... (i)$For the second case: $e(1.6) = \frac{hc}{2\lambda} - \phi_0 \quad ... (ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{4.8}{1.6} = \frac{\frac{hc}{\lambda} - \phi_0}{\frac{hc}{2\lambda} - \phi_0}$$3 = \frac{\frac{hc}{\lambda} - \phi_0}{\frac{hc}{2\lambda} - \phi_0}$$3 \left( \frac{hc}{2\lambda} - \phi_0 \right) = \frac{hc}{\lambda} - \phi_0$$\frac{3hc}{2\lambda} - 3\phi_0 = \frac{hc}{\lambda} - \phi_0$$\frac{3hc}{2\lambda} - \frac{hc}{\lambda} = 2\phi_0$$\frac{hc}{2\lambda} = 2\phi_0 \implies \phi_0 = \frac{hc}{4\lambda}$Since $\phi_0 = \frac{hc}{\lambda_0}$,we have $\frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$.Therefore,the threshold wavelength $\lambda_0 = 4\lambda$.

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. The threshold wavelength for the surface is:

Similar Questions

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity:

The silicate having one monovalent corner oxygen atom in each tetrahedron unit is:

An integrating factor of the equation $(1+y+x^2 y) dx+(x+x^3) dy=0$ is

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides of the triangle are in

There are three lumps of a given radioactive substance. Their activity is in the ratio of $1 : 2 : 3$ now. What will be the ratio of their activities at any further date?

Vedclass Test Series

Exam Paper Generator

Online Exam Module