Question

The ratio of the de-Broglie wavelength of a deuterium atom to that of $\alpha -partical$ when then the velocity of former is five times greater than that of later is:

• A. 4
• B. 10.2
• C. 2
• D. 0.4

Answer 1

The correct option is D 0.4

The correct option is D We have,

Deuterium atom and α−partical According to the De-Broglie wavelength

$\lambda =\frac{h}{mv}$

Where $\lambda$ is the wavelength, h is Planks constant m is mass and v

is the velocity of the particle.

For deuterium,

mass of deuterium $=2amu$

The speed of Deuterium $=v$

So, wavelength of deuterium $=\frac{h}{2v}....1$

For alpha particle,

Mass of alpha particle $4amu$

speed of the alpha particle $=\frac{v}{5}$

So,

Wavelength of the alpha particle

$\frac{h}{4}\times \frac{v}{5}$

$=\frac{5h}{4v}....2$

NOw, the ratio of the wavelength of the deuterium to alpha particle

$=\frac{h}{2v}\frac{5h}{4v}$

$\frac{4}{10}=0.4$