Question

A proton is accelerated through a potential difference of $V$ volts. The de Broglie's wavelength associated with it is:

• A. $\sqrt{\frac{150}{V}}\stackrel{\circ }{A}$
• B. $\frac{0.287}{\sqrt{V}}\stackrel{\circ }{A}$
• C. $\frac{0.202}{\sqrt{V}}\stackrel{\circ }{A}$
• D. $\frac{0.101}{\sqrt{V}}\stackrel{\circ }{A}$

Answer 1

The correct option is B $\frac{0.287}{\sqrt{V}}\stackrel{\circ }{A}$

Given:
Applied potential difference = $V$ volts

We know that,
$\text{Mass of}proton\left(m\right)=1.67\times {10}^{-27}kg$
$\text{Charge on}proton=e(e=1.6\times {10}^{-19}coulomb)$

Hence, the final kinetic energy of the proton, $\left(KE\right)=V\times e$

de Broglie wavelength$\left(\lambda \right)$ in terms of kinetic energy (KE) can be given as:
$\lambda =\frac{h}{\sqrt{2KEm}}$

Since, $KE=V\times e\Rightarrow \lambda =\frac{h}{\sqrt{2\times V\times e\times m}}=\frac{6.63\times {10}^{-34}}{\sqrt{2\times V\times 1.6\times {10}^{-19}\times 1.67\times {10}^{-27}}}=\frac{2.87\times {10}^{-11}}{\sqrt{V}}m=\frac{0.287}{\sqrt{V}}\stackrel{\circ }{A}$