Question: Q. 3. A proton, a neutron, an electron and an $\alpha$-particle have same energy. Then, their de-Broglie wavelengths compare as (a) $\lambda_{p}=\lambda_{n}>\lambda_{e}>\lambda_{\alpha}$. (b) $\lambda_{\alpha}<\lambda_{p}=\lambda_{n}<\lambda_{e}$. (c) $\lambda_{e}<\lambda_{p}=\lambda_{n}>\lambda_{\alpha}$. (d) $\lambda_{e}=\lambda_{p}=\lambda_{n}=\lambda_{\alpha}$
[NCERT Exemplar]
Show Answer
Solution:
Ans. Correct option : (b) Explanation :
According to de-Broglie, a moving material particle sometimes acts as a wave and sometimes as a particle.
De- Broglie wavelength, $\lambda_{d}=\frac{h}{p}$
$$ \begin{aligned} E_{p} & =E_{n}=E_{e}=E_{\alpha} \ K . E . & =K=\frac{1}{2} m v^{2} \ 2 K & =m v^{2} \ 2 K m & =m^{2} v^{2} \ 2 m K & =p^{2} \ \sqrt{2 m K} & =p \ \therefore \quad \lambda_{d} & =\frac{h}{p} \ \lambda_{d} & =\frac{h}{\sqrt{2 m K}} \ \text { or } \lambda_{d} & =\frac{h}{\sqrt{2 m K}}[\text { as } h \text { and } E \text { (K.E.) is constt. }] \ \therefore \lambda^{2} & \frac{1}{\sqrt{m}} \ m_{a}>m_{p} & =m_{n}>m_{e} \ \therefore \lambda_{a} & <\lambda_{p}=\lambda_{n}<\lambda_{e} \end{aligned} $$
