A metal surface is illuminated by light of tw | vedclass

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

Question

(A) The energy of the incident photons is given by $E = \frac{hc}{\lambda}$.For $\lambda_1 = 248 \ nm$,$E_1 = \frac{1240}{248} = 5 \ eV$.For $\lambda_

Vedclass · Accepted Answer

$3.7$

Vedclass · Answer

(A) The energy of the incident photons is given by $E = \frac{hc}{\lambda}$.For $\lambda_1 = 248 \ nm$,$E_1 = \frac{1240}{248} = 5 \ eV$.For $\lambda_2 = 310 \ nm$,$E_2 = \frac{1240}{310} = 4 \ eV$.According to Einstein's photoelectric equation,the maximum kinetic energy is $K.E. = E - W$,where $W$ is the work function.Thus,$K.E._1 = 5 - W$ and $K.E._2 = 4 - W$.Since $K.E. = \frac{1}{2}mv^2$,the ratio of kinetic energies is $\frac{K.E._1}{K.E._2} = \left(\frac{u_1}{u_2}\right)^2 = \left(\frac{2}{1}\right)^2 = 4$.Therefore,$\frac{5 - W}{4 - W} = 4$.$5 - W = 4(4 - W) = 16 - 4W$.$3W = 11$.$W = \frac{11}{3} \approx 3.67 \ eV \approx 3.7 \ eV$.

Similar Questions

Light of wavelength $1824 \mathring A$,incident on the surface of a metal,produces photo-electrons with maximum energy $5.3 \ eV$. When light of wavelength $1216 \mathring A$ is used,the maximum energy of photoelectrons is $8.7 \ eV$. The work function of the metal surface is .............. $eV$.

The work function of a metal is $3 \ eV$. The color of the visible light that is required to cause emission of photoelectrons is

When a photosensitive surface is irradiated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively. The work function of the photosensitive surface is:

The work function of a material is $4.0 \ eV$. The maximum wavelength of light that can cause the emission of photoelectrons from the material is ............ $nm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module