A Proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is

Question

A Proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is (A) 2 (B) √2 (C) 2√2 (D) (1/2√2)

Tardigrade Admin · Accepted Answer

Key Point The de-Broglie wavelength of a particle of mass

m

and moving with velocity

v

is given by

λ = \frac{h}{m v} (∵ p = m v)

de-Broglie wavelength of a proton of mass

m_{1}

and kinetic energy

k

is given by

λ_{1} = \frac{h}{2 m _{1} k} (∵ p = 2 mk)

= \frac{h}{2 m _{1} q v} \dots (i) [∵ k = q V]

For an alpha particle mass

m_{2}

carrying charge

q_{0}

is accelerated through potential

V

, then

λ_{2} = \frac{h}{2 m _{2} q _{0} V}

∵

For

α

-particle

(_{2}^{4} He)

∴ q_{0} = 2 q

and

m_{2} = 4 m_{1}

∴ λ_{2} = \frac{h}{2 \times 4 m _{1} \times 2 q \times V}

...(ii)
The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get

\frac{λ _{1}}{λ _{2}} = \frac{h}{2 m _{1} q V} \times \frac{2 \times m _{1} \times 4 \times 2 q V}{h} = \frac{4}{2} \times \frac{2}{2}

= 22

Q. A Proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is

$2$

$2$

$22$

$\frac{1}{2 2}$

Q. A Proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is

2

2​

22​

22​1​

$2$

$2$

$22$

$\frac{1}{2 2}$