Question

If $E_1$,$E_2$ and $E_3$ are the respective kinetic energies of an electron, an alpha particle and a proton each having the same de Brogli wavelength then

• A. $E_1$ > $E_3$ > $E_2$
• B. $E_2$ > $E_3$ > $E_1$
• C. $E_1$ > $E_2$ > $E_3$
• D. $E_1$ = $E_2$ = $E_3$

Answer 1

The correct option is A $E_1$ > $E_3$ > $E_2$

Kinetic energy is $\frac{p^2}{2m}$ ..................................... $1$

where $p$ is momentum and $m$ is mass of the particle.

We also know that the De Broglie wavelength is $\lambda =h/p$ ............. $2$

where $h$ is Planck's constant and $p$ is momentum.

As it is given that $\lambda$ is same for all the three particles so the all the particles should have same value of momentum $p$. ( See equation 2)

Now from the relation 1 we get for same value of $p$ we have kinetic energy inversely proportional to $mass$ $m$.

i.e. $K.E$ for electron (lowest mass) will be highest and $K.E.$ for alpha particle (highest mass) as lowest.

So $E_1>E_3>E_2$ .