Question

The de Broglie wavelength of a neutron corresponding to root mean square speedat ${927}^{\circ }C\text{is}\lambda$. What will be the de Broglie wavelength of the root mean square speed at ${27}^{\circ }C$ ?

• A. $\frac{\lambda }{2}$
• B. $\lambda$
• C. $2\lambda$
• D. $4\lambda$

Answer 1

The correct option is C $2\lambda$

Root mean square speed is given by $V_m=\sqrt{\frac{3RT}{M_w}}\Rightarrow V_m\propto \sqrt{T}$
Ratio of rms speed at different temperatures is
$\frac{V_2}{V_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{1200}{300}}=2$

$V_2=2V_1$

We know that
$V\propto \lambda$
As $V$ is doubled, $\lambda$ will also be doubled.