Electrons with de-Broglie wavelength lambda f | vedclass

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

Question

(A) The de-Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.Squaring bot

Vedclass · Accepted Answer

$\lambda_0 = \frac{2mc\lambda^2}{h}$

Vedclass · Answer

(A) The de-Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.Squaring both sides,we get $\lambda^2 = \frac{h^2}{2meV}$,which implies $eV = \frac{h^2}{2m\lambda^2}$.The cut-off wavelength $\lambda_0$ of the emitted $X$-rays corresponds to the maximum energy of the photon,which is equal to the kinetic energy of the incident electron: $E = eV = \frac{hc}{\lambda_0}$.Equating the two expressions for $eV$,we have $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$.Solving for $\lambda_0$,we get $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

Electrons with de-Broglie wavelength $\lambda$ fall on the target of an $X$-ray tube. The cut-off wavelength of the emitted $X$-ray is:

Similar Questions

In an $LCR$ circuit,the capacitance is changed from $C$ to $2C$. For the resonant frequency to remain unchanged,the inductance should be changed from $L$ to:

The equation of trajectory of a projectile is $y=10 x-\left(\frac{5}{9}\right) x^2$. If we assume $g=10 \ m/s^2$,the range of the projectile (in metres) is:

Where are the Cardamom Hills located?

Charges of $+\frac{10}{3} \times 10^{-9} \text{ C}$ are placed at each of the four corners of a square of side $8 \text{ cm}$. The potential at the intersection of the diagonals is:

Which of the following is an example of a biofertilizer?

Vedclass Test Series

Exam Paper Generator

Online Exam Module