JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 3)
An $$\alpha$$ particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are $$\lambda$$$$\alpha$$ and $$\lambda$$p respectively. The ratio $${{{{\lambda _p}} \over {{\lambda _\alpha }}}}$$ is :
8
2.8
7.8
3.8
Explanation
We know,
$$qv = {{{p^2}} \over {2m}}$$
$$ \Rightarrow p = \sqrt {2mqv} $$
$$ \therefore $$ $$\lambda = {h \over {\sqrt {2mqv} }}$$
$$ \therefore $$ $${{{\lambda _p}} \over {{\lambda _\alpha }}} = {{\sqrt {2{m_\alpha }{q_\alpha }v} } \over {\sqrt {2{m_p}{q_p}v} }}$$
$$ = {{\sqrt {2(4m)(2e)} } \over {\sqrt {2me} }}$$
$$ = \sqrt 8 $$
$$ = 2\sqrt 2 $$
$$qv = {{{p^2}} \over {2m}}$$
$$ \Rightarrow p = \sqrt {2mqv} $$
$$ \therefore $$ $$\lambda = {h \over {\sqrt {2mqv} }}$$
$$ \therefore $$ $${{{\lambda _p}} \over {{\lambda _\alpha }}} = {{\sqrt {2{m_\alpha }{q_\alpha }v} } \over {\sqrt {2{m_p}{q_p}v} }}$$
$$ = {{\sqrt {2(4m)(2e)} } \over {\sqrt {2me} }}$$
$$ = \sqrt 8 $$
$$ = 2\sqrt 2 $$
Comments (0)
