JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 10)
An electron having de-Broglie wavelength $$\lambda$$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is :
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$${{2{m^2}{c^2}{\lambda ^2}} \over {{h^2}}}$$
$${{2mc{\lambda ^2}} \over h}$$
$${{hc} \over {mc}}$$
Explanation
$$\lambda = {h \over {mv}}$$
kinetic energy, $${{{P^2}} \over {2m}} = {{{h^2}} \over {2m{\lambda ^2}}} = {{hc} \over {{\lambda _c}}}$$
$${\lambda _c} = {{2mc{\lambda ^2}} \over h}$$
kinetic energy, $${{{P^2}} \over {2m}} = {{{h^2}} \over {2m{\lambda ^2}}} = {{hc} \over {{\lambda _c}}}$$
$${\lambda _c} = {{2mc{\lambda ^2}} \over h}$$
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