Question

An electron with initial kinetic energy of $100\text{eV}$ is accelerated through a potential difference of $50\text{V}.$ The de-Broglie wavelength of the electron is-

• A. $1\mathring{\text{A}}$
• B. $\sqrt{1.5}\mathring{\text{A}}$
• C. $\sqrt{3}\mathring{\text{A}}$
• D. $12.27\mathring{\text{A}}$

Answer 1

Answer: (A)

We are given, the initial kinetic energy electron = 100 eV

Then electron is accelerated through a potential difference, so kinetic energy gained during this is given by formula

K' = qV ; where q is charge on particle and V is potential difference in which it moves.

Here, K' = eV ; e is charge on an electron

Potential difference by which it is accelerated is 50V

So gain of kinetic energy is K' =e(50V)=50eV

Total kinetic energy of electron after acceleration is equal to K =50 eV +100 eV =150 eV

In terms of joules K = $150\times 1.6\times {10}^{-19}J$

The de-Broglie wavelength can be calculated from the following formula:

$\lambda =\frac{h}{\sqrt{2\times m\times K}}$ ; $\lambda$ is wavelength , m is mass of particle, K is kinetic energy, h is planck's constant

Here, m is mass of electron = $9.1\times {10}^{-31}$kg , planck's constant =$6.63\times {10}^{-34}$ Js , K=$150\times 1.6\times {10}^{-19}J$

Put in formula we get:

$\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 150\times 1.6\times {10}^{-19}}}$

$\lambda =\frac{6.63\times {10}^{-34}}{66.09\times {10}^{-25}}\approx 1\times {10}^{-10}m$ = 1$A^o$

Hence, the de-Broglie wavelength is 1$A^o$ .