Question

Graph shows the variation of de-Broglie wavelength `(lambda)` versus `1/sqrt"V"`, where 'V' is the accelerating potential for four particles carrying same charge but of masses m₁ , m₂, m₃, m₄. Which particle has a smaller mass?

Options

• m₁

• m₃

• m₄

• m₂

Answer 1

m₄

Explanation:

`"de Broglie wavelength"  lambda = "h"/"p"`

`therefore lambda = "h"/sqrt(2"mqv")`

`therefore lambda sqrt"v" = "h"/sqrt(2"mq")` or `lambda/((1/sqrt"v")) = 1/sqrt(2"mq")`

`therefore  "Slope of the graph" = 1/sqrt(2"mq")`

The slope will be maximum for minimum mass.

∴ m₄ is minimum.