JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 8)
The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (mp = 1.00727u me = 0.00055u)
41.4 : 1
(1860)2 : 1
1860 : 1
43 : 1
Explanation
$${\lambda _e} = {{12.27} \over {\sqrt V }}\mathop A\limits^o $$
$${\lambda _p} = {{0.286} \over {\sqrt V }}\mathop A\limits^o $$
$${{{\lambda _e}} \over {{\lambda _p}}} = {{12.27} \over {0.286}} = 43$$
$${\lambda _p} = {{0.286} \over {\sqrt V }}\mathop A\limits^o $$
$${{{\lambda _e}} \over {{\lambda _p}}} = {{12.27} \over {0.286}} = 43$$
Comments (0)
