An electron beam is accelerated by a potentia | vedclass

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

Question

(D) The energy of the electron is given by $E = eV$,where $e$ is the charge of the electron and $V$ is the accelerating potential difference.This ener

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) The energy of the electron is given by $E = eV$,where $e$ is the charge of the electron and $V$ is the accelerating potential difference.This energy is converted into the energy of the emitted $X$-ray photon,$E = \frac{hc}{\lambda}$.Equating the two,we get $eV = \frac{hc}{\lambda}$,which implies $\lambda = \frac{hc}{eV}$.Since $h$,$c$,and $e$ are constants,the minimum wavelength $\lambda_{min}$ is inversely proportional to the accelerating potential $V$ $(\lambda_{min} \propto \frac{1}{V})$.Therefore,to obtain the shortest wavelength,the potential difference $V$ must be the maximum among the given options.Comparing the options $10 \ kV, 20 \ kV, 30 \ kV,$ and $40 \ kV$,the maximum value is $40 \ kV$.

An electron beam is accelerated by a potential difference of $V$ and strikes a metal target to produce $X$-rays. For which of the following values of $V$ (in $kV$) will the $X$-ray wavelength be the shortest?

Similar Questions

How are $X$-rays produced in the laboratory?

The energy of an electromagnetic radiation is $14.4 \, KeV$. To which part of the electromagnetic spectrum does it belong?

The continuous $X-$ray spectrum produced by an $X-$ray machine at constant voltage has

The minimum wavelength of $X$-rays produced by electrons accelerated by a potential difference of $V$ volts is equal to:

$A$ metal block is exposed to beams of $X$-rays of different wavelengths. $X$-rays of which wavelength penetrate most?

Vedclass Test Series

Exam Paper Generator

Online Exam Module