Question

If kinetic energy of a proton is increased nine times the wavelength of the de broglie wave associated with it would become

Answer 1

From de-Broglie formula:
λ = h / √2m K.E
For a proton : K.E = k → 9k ⇒ k ' = 9k
m' = m
λ' = h / √ 2 m' k'  = h / √ 2m 9k  =  λ / 3

So new wavelength decreases by 3 times .

Answer 2

Answer: De-Broglie's wavelength is reduced to $\frac{1}{3}$ the original value.

Explanation:

The wavelength associated with De-Broglie is given as:

$\lambda=\frac{h}{\sqrt{2mE}}$     ......(1)

where,

$\lambda$ = De-Broglie's wavelength

h = Planck's constant

m = Mass of the particle

E = Kinetic energy of the particle

It is given that:

$E'=9E$

$\lambda'=\frac{h}{\sqrt{2mK'}}$         ......(2)

Taking ratio of  2 and 1, we get:

$\frac{\lambda'}{\lambda}=\left(\frac{\frac{h}{\sqrt{2m9E}}}{\frac{h}{\sqrt{2mE}}}\right)\\\\\frac{(\lambda')^2}{(\lambda)^2}=\frac{1}{9}\\\\\lambda'=\frac{\lambda}{3}$

Hence, de-Broglie's wavelength is reduced to $\frac{1}{3}$ the original value.