If v_0 is the threshold frequency of a metal | vedclass

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

Question

(A) According to the de Broglie wavelength equation,$\lambda = \frac{h}{p} = \frac{h}{mv}$.Since kinetic energy $KE = \frac{1}{2}mv^2$,we have $v = \s

Vedclass · Accepted Answer

$\lambda \propto \frac{1}{\sqrt{v-v_0}}$

Vedclass · Answer

(A) According to the de Broglie wavelength equation,$\lambda = \frac{h}{p} = \frac{h}{mv}$.Since kinetic energy $KE = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2 	imes KE}{m}}$.Substituting this into the wavelength equation: $\lambda = \frac{h}{\sqrt{2m 	imes KE}}$.From the photoelectric effect equation,$KE = hv - hv_0 = h(v - v_0)$.Substituting $KE$ into the wavelength expression: $\lambda = \frac{h}{\sqrt{2mh(v - v_0)}} = \sqrt{\frac{h^2}{2mh(v - v_0)}} = \sqrt{\frac{h}{2m(v - v_0)}}$.Therefore,$\lambda \propto \frac{1}{\sqrt{v - v_0}}$ or $\lambda \propto (v - v_0)^{-\frac{1}{2}}$.

If $v_0$ is the threshold frequency of a metal $X$,the correct relation between de Broglie wavelength $(\lambda)$ associated with the photoelectron and the frequency $(v)$ of the incident radiation is:

Similar Questions

The de-Broglie's wavelength of an electron in the $4^{th}$ orbit is . . . . . . . . $\pi a_0$. ($a_0 =$ Bohr's radius)

The de Broglie wavelength of a particle $X$ moving with a certain velocity is $1 \ \mathring{A}$. If the mass of $Y$ is $25\%$ of $X$ and its velocity is $75\%$ of $X$,then the de Broglie wavelength of $Y$ will be ........... $\mathring{A}$.

Can the wave nature of ordinary objects be detected? Why?

The de Broglie wavelength of a tennis ball of mass $60 \ g$ moving with a velocity of $10 \ m/s$ is approximately ........

If the kinetic energy of an electron is $1.25 \times 10^{-28} \ J$,then what will be the de-Broglie wavelength of the electron? (Mass of electron $= 9 \times 10^{-31} \ kg$,$h = 6.6 \times 10^{-34} \ Js$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module