A photoelectric surface is illuminated succes | vedclass

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

Question

(B) Let $\phi_{0}$ be the work function of the surface of the material. According to Einstein's photoelectric equation,the maximum kinetic energy of t

Vedclass · Accepted Answer

$\frac{hc}{2\lambda}$

Vedclass · Answer

(B) Let $\phi_{0}$ be the work function of the surface of the material. According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted photoelectrons in the first case is $K_{\max 1} = \frac{hc}{\lambda} - \phi_{0}$.In the second case,the wavelength is $\lambda/2$,so the maximum kinetic energy is $K_{\max 2} = \frac{hc}{\lambda/2} - \phi_{0} = \frac{2hc}{\lambda} - \phi_{0}$.Given that $K_{\max 2} = 3 K_{\max 1}$,we substitute the expressions:$\frac{2hc}{\lambda} - \phi_{0} = 3 \left( \frac{hc}{\lambda} - \phi_{0} \right)$.Expanding the equation: $\frac{2hc}{\lambda} - \phi_{0} = \frac{3hc}{\lambda} - 3\phi_{0}$.Rearranging the terms: $3\phi_{0} - \phi_{0} = \frac{3hc}{\lambda} - \frac{2hc}{\lambda}$.$2\phi_{0} = \frac{hc}{\lambda}$.Therefore,the work function is $\phi_{0} = \frac{hc}{2\lambda}$.

Similar Questions

Light of incident frequency $2$ times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{3}\right)^{\text{rd}}$ and intensity is doubled,then the photoelectric current will

The work function of caesium is $2.14\, eV$. Find the wavelength of the incident light if the photo current is brought to zero by a stopping potential of $0.60\, V$. (Result in $nm$)

When a metal is irradiated by light having wavelength $\lambda$ $(\lambda < \lambda_0)$,all the photoelectrons emitted are bent in a circle of radius $r$ by a magnetic field of flux density $B_0$. Find $\frac{1}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.

In a photoelectric effect experiment,the threshold wavelength of incident light is $380 \, nm$. If the wavelength of the incident light is $260 \, nm$,the maximum kinetic energy of emitted electrons will be: .............. $eV$. Given $E \text{ (in } eV) = \frac{1237}{\lambda \text{ (in } nm)}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module