The threshold frequency of a photosensitive m | vedclass

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

Question

(C) The frequency of the $H_{\alpha}$ line of the Balmer series is given by $
u_0 = Rc(\frac{1}{2^2} - \frac{1}{3^2}) = Rc(\frac{1}{4} - \frac{1}{9})

Vedclass · Accepted Answer

$\frac{7 Rhc}{144}$

Vedclass · Answer

(C) The frequency of the $H_{\alpha}$ line of the Balmer series is given by $
u_0 = Rc(\frac{1}{2^2} - \frac{1}{3^2}) = Rc(\frac{1}{4} - \frac{1}{9}) = \frac{5Rc}{36}$. This is the threshold frequency $
u_0$ of the material.The frequency of the $H_{\beta}$ line of the Balmer series is given by $
u = Rc(\frac{1}{2^2} - \frac{1}{4^2}) = Rc(\frac{1}{4} - \frac{1}{16}) = \frac{3Rc}{16}$.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h
u - h
u_0$.Substituting the values:$K_{max} = h(\frac{3Rc}{16}) - h(\frac{5Rc}{36})$$K_{max} = Rhc(\frac{3}{16} - \frac{5}{36})$$K_{max} = Rhc(\frac{27 - 20}{144}) = \frac{7 Rhc}{144}$.

Similar Questions

When a photon of wavelength ${\lambda _1}$ is incident on a metal surface,the speed of the ejected fastest electron is $v_1 \ m/s$. If a photon of wavelength ${\lambda _2}$ is incident on the same metal,the speed of the ejected fastest electron is $v_2 \ m/s$. The value of $v_2^2 - v_1^2$ will be

When ultraviolet rays are incident on a metal plate, the photoelectric effect does not occur. It occurs by the incidence of:

The work function of Cesium is $2.27 eV$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 nm$ wavelength is

The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module