Question

In which of the following situations, the heavier of the two particles has smaller de Broglie wavelength? The two particles

• A. Move with the same speed
• B. Move with the same linear momentum
• C. Move with the same kinetic energy
• D. Have fallen through the same height

Answer 1

The correct option is C

Move with the same kinetic energy

Take 2 particle with $m_{1}and{m}_2$

$m_2>m_1$

Now,

$K{E}_1=K{E}_2$ $\Rightarrow$ $\frac{1}{2}$ $m_1$ ${v_1}^2=\frac{1}{2}$ $m_2$ ${v_2}^2$

Because $m_2$ > $m_1$ $\Rightarrow$ $v_1>v_2$

But $m_2{v}_2>m_1{v}_1$ [Because v is squared]

Therefore, for De-Brogile wavelength,

${\lambda }_1=\frac{h}{m_1{v}_1}and{\lambda }_2$ = $\frac{h}{m_2{v}_2}$

Because $m_2{v}_2>m_1{v}_1$

$\Rightarrow$ ${\lambda }_1$ > ${\lambda }_2$

Hence, it is possible when kinetic energy is same.