Question

If two particles of equal masses have kinetic energy in the ratio 1 : 4. The ratio of their de Broglie wavelength will be

Answer 1

Explanation:

2

1

mv

2

=E

mv=

2Em

De-broglie wavelength

λ=

mv

h

λ

m

=

2E×m

h

λ

2m

=

2×E×2m

h

λ

2m

λ

m

=

1

2

Answer 2

Explanation:

Given :

Ratio of Kinetic Energies of two particles having same mass is 1 : 4

To Find :

Ratio of their De-Broglie wavelength

Solution :

The relation between Kinetic energy and de-broglie wavelength is given by ,

{\boxed {\rm{ \lambda = \frac{h}{ \sqrt{2mKE} } }}}

λ=

2mKE

h

Where ,

λ is wavelength

h is planck's constant

m is mass

KE is Kinetic Energy

We are given that the particle's masses are equal so let the masses be 'm' and 'm'

Let ,

Kinetic energy of the first particle be (KE)₁

Kinetic energy of second particle be (KE)₂

Debriglie wavelength of first particle be λ₁

De-broglie wavelength of second particke be λ₂

We have ,

(KE)₁ : (KE)₂ = 1 : 4

Let

(KE)₁ = x , (KE)₂ = 4x

Now ,

\begin{gathered} : \implies \sf \: \dfrac{ \lambda_1}{ \lambda_2} = \dfrac{ \frac{h}{ \sqrt{2m(KE)_1} } }{ \frac{h}{ \sqrt{2m(KE)_2} }} \\ \\ : \implies \sf \: \frac{\lambda_1}{\lambda_2} = \dfrac{ \frac{1}{ \sqrt{2m(x)} } }{ \frac{1}{ \sqrt{2m(4x)} } } \\ \\ : \implies \sf \: \dfrac{ \lambda_1}{\lambda_2} = \frac{ \sqrt{2m(4x)} }{ \sqrt{2m(x)} } \\ \\ : \implies \sf \: \frac{\lambda_1}{ \lambda_2} = \frac{ \sqrt{4x} }{ \sqrt{x} } \\ \\ : \implies \sf \: \frac{ \lambda_1}{ \lambda_2} = \frac{2 \sqrt{x} }{ \sqrt{x} } \\ \\ : \implies \sf \: \frac{ \lambda_1}{ \lambda_2} = \frac{2}{1} \\ \\ : \implies \sf {\boxed {\sf{ \blue {\lambda_1 : \lambda_2 = 2 : 1}}}}\end{gathered}

:⟹ sum is in picture

∴ The ratio of De-broglie wavelength of the given particles is 2 : 1