JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 21)
If the de Broglie wavelength of an electron is equal to the 10–3 times the wavelength of a photon of frequency 6 $$ \times $$ 1014 Hz, then the speed of electron is equal to : (Speed of light = 3 $$ \times $$ 108 m/s, Planck's constant = 6.63 $$ \times $$
10–34 J.s, Mass of electron = 9.1 $$ \times $$ 10–31 kg)
1.7 $$ \times $$ 106 m/s
1.45 $$ \times $$ 106 m/s
1.1 $$ \times $$ 106 m/s
1.8 $$ \times $$ 106 m/s
Explanation
$${h \over {mv}} = {10^{ - 3}}\left( {{{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}}} \right)$$
v $$ = {{6.63 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^5}}}$$
v $$ = 1.45 \times {10^6}$$ m/s
v $$ = {{6.63 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^5}}}$$
v $$ = 1.45 \times {10^6}$$ m/s
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