Energy required to remove an electron from an | vedclass

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

Question

(A) Given: Work function $\phi_0 = 4.2 	ext{ eV}$,Wavelength $\lambda = 2000 	ext{ Å}$.Energy of incident photon $E = \frac{12400}{\lambda(	ext{in

Vedclass · Accepted Answer

$8.4 	imes 10^5 	ext{ m/s}$

Vedclass · Answer

(A) Given: Work function $\phi_0 = 4.2 	ext{ eV}$,Wavelength $\lambda = 2000 	ext{ Å}$.Energy of incident photon $E = \frac{12400}{\lambda(	ext{in Å})} 	ext{ eV} = \frac{12400}{2000} = 6.2 	ext{ eV}$.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max} = E - \phi_0$.$K_{\max} = 6.2 	ext{ eV} - 4.2 	ext{ eV} = 2 	ext{ eV}$.Converting kinetic energy to Joules: $K_{\max} = 2 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-19} 	ext{ J}$.Using $K_{\max} = \frac{1}{2} m v_{\max}^2$,where mass of electron $m = 9.1 	imes 10^{-31} 	ext{ kg}$.$v_{\max}^2 = \frac{2 	imes K_{\max}}{m} = \frac{2 	imes 3.2 	imes 10^{-19}}{9.1 	imes 10^{-31}} \approx 0.703 	imes 10^{12} 	ext{ m}^2/	ext{s}^2$.$v_{\max} = \sqrt{0.703 	imes 10^{12}} \approx 8.38 	imes 10^5 	ext{ m/s} \approx 8.4 	imes 10^5 	ext{ m/s}$.

Energy required to remove an electron from an aluminum surface is $4.2 \text{ eV}$. If light of wavelength $2000 \text{ Å}$ falls on the surface,the velocity of the fastest electron ejected from the surface will be:

Similar Questions

The work function of a metal is $2.1 \text{ eV}$. Which of the following wavelengths will be able to emit photoelectrons from its surface?

Ultraviolet radiations of $6.2 \, eV$ fall on an aluminium surface (work function $4.2 \, eV$). The kinetic energy in joules of the fastest electron emitted is approximately:

$A$ monochromatic neon lamp with a wavelength of $670.5\, nm$ illuminates a photo-sensitive material which has a stopping voltage of $0.48\, V$. What will be the stopping voltage if the source light is changed with another source of wavelength $474.6\, nm$? (In $V$)

In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)

Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is $4.2 \ eV$. In order to reduce the photo-current to zero,the voltage of the anode relative to the cathode must be made ........... $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module