Question

The energy of a photon whose de-Broglie wavelength is equal to the wavelength of an electron accelerated through a potential difference of $125\text{V}$ is near to
[Use $hc=12400\text{eV}\mathring{\text{A}}$]

• A. $11.3\text{eV}$
• B. $11.3\text{keV}$
• C. $125\text{eV}$
• D. $1250\text{eV}$

Answer 1

The correct option is B $11.3\text{keV}$

Given, $V=125\text{V}$

de-Broglie wavelength associated with an electron accelerated under a potential difference $V$ is,

$\lambda =\frac{12.27}{\sqrt{V}}\mathring{A}=\frac{12.27}{\sqrt{125}}\mathring{A}=1.097\mathring{A}$

Now, energy of photon,

$E=\frac{hc}{\lambda }=\frac{12400\text{eV}\left(\mathring{A}\right)}{1.097\mathring{A}}\approx 11.3\times {10}^3\text{eV}=11.3\text{keV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.

Why this question? Key concept: de-Broglie wavelength associated with an electron accelerated under a potential difference $V$ is, $\lambda =\frac{12.27}{\sqrt{V}}\mathring{A}$