JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
The ratio of de-Broglie wavelength of an $$\alpha$$-particle and a proton accelerated from rest by the same potential is $$\frac{1}{\sqrt{m}}$$, the value of $$m$$ is:
Solution :
De-Broglie wavelength is given by :
$$\lambda = \frac{h}{\sqrt{2mqV}}$$
For α-particle :
Mass,
$$m_\alpha = 4m_p$$
Charge,
$$q_\alpha = 2e$$
Therefore,
$$\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)V}}$$
$$= \frac{h}{\sqrt{16m_peV}}$$
$$= \frac{h}{4\sqrt{m_peV}}$$
For proton :
$$\lambda_p = \frac{h}{\sqrt{2m_peV}}$$
Comparing with :
$$\frac{1}{\sqrt{m}}$$
we get,
$$m = 8$$
Final Answer :
$$8$$
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
