The masses of two particles having same kinetic energy are in the ratio of 2 ∶ 1. Their de Broglie wavelengths are in the ratio

  1. 2 ∶ 1
  2. 1 ∶ 2
  3. √ 2 : 1
  4.  1: √ 2

Answer (Detailed Solution Below)

Option 4 :  1: √ 2
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Detailed Solution

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Concept:

The de-Broglie wavelength is given by:

\(\lambda = \frac{h}{{mv}} = \frac{h}{p}\)

m is the mass of the particle

h is the plank's constant

v is the velocity of the particle

p = momentum.

As can be seen from the relation "The de-Broglie wavelength is inversely proportional to the mass of the particle and its velocity but is independent of the nature of the particle."

\(\Rightarrow p = \frac{h}{\lambda }\)

The Kinetic energy in terms of momentum is given as:

\(K.E.\; = \frac{{{p^2}}}{{2m}}\)

Using the expression of momentum (p) in K.E.

\(K.E. = \frac{{{h^2}}}{{2m\;{\lambda ^2}}}\)

Calculation:

Given,

K.E. (1st object) = K.E. (2nd Object)

The ratio of their masses = \(\frac{{{m_1}}}{{{m_2}}} = \frac{2}{1}\)

We know that,

\(K.E. = \frac{{{h^2}}}{{2m\;{\lambda ^2}}}\)

\( \Rightarrow {\lambda ^2} = \frac{{{h^2}}}{{2m\;\left( {K.E} \right)}}\)

Since (K.E)1 = (K.E)2

\(\frac{{{{\left( {{\lambda _1}} \right)}^2}}}{{{{\left( {{\lambda _2}} \right)}^2}}} = \frac{{\left( {\frac{{{h^2}}}{{2{m_1}\;{{\left( {K.E} \right)}_1}}}} \right)}}{{\left( {\frac{{{h^2}}}{{2{m_2}\;{{\left( {K.E} \right)}_2}}}} \right)}} = \frac{{{m_2}}}{{{m_1}}} = \frac{1}{2}\)

\(\frac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\frac{1}{2}} = \frac{1}{{\sqrt 2 }}\)

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