JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 7)
A particle is moving 5 times as fast as an
electron. The ratio of the de-Broglie wavelength
of the particle to that of the electron is 1.878 $$ \times $$
10–4. The mass of the particle is close to
1.2 $$ \times $$ 10–28 kg
9.1 $$ \times $$ 10–31 kg
4.8 $$ \times $$ 10–27 kg
9.7 $$ \times $$ 10–28 kg
Explanation
Let mass of particle = m
Let speed of e– = V
$$ \therefore $$ speed of particle = 5V
de-broglie wavelength $$\lambda $$d = $${h \over P} = {h \over {mv}}$$
$$ \therefore $$ ($$\lambda $$d)P = $${h \over {m\left( {5V} \right)}}$$ ....(1)
and ($$\lambda $$d)e = $${h \over {{m_e}\left( V \right)}}$$ ....(1)
According to question
$${{{{\left( {{\lambda _d}} \right)}_P}} \over {{{\left( {{\lambda _d}} \right)}_e}}} = {{{m_e}} \over {5m}}$$ = 1.878 $$ \times $$ 10–4
$$ \Rightarrow $$ m = $${{{m_e}} \over {5 \times 1.874 \times {{10}^{ - 4}}}}$$
= $${{9.1 \times {{10}^{ - 31}}} \over {5 \times 1.874 \times {{10}^{ - 4}}}}$$
= 9.7 $$ \times $$ 10–28 kg
Let speed of e– = V
$$ \therefore $$ speed of particle = 5V
de-broglie wavelength $$\lambda $$d = $${h \over P} = {h \over {mv}}$$
$$ \therefore $$ ($$\lambda $$d)P = $${h \over {m\left( {5V} \right)}}$$ ....(1)
and ($$\lambda $$d)e = $${h \over {{m_e}\left( V \right)}}$$ ....(1)
According to question
$${{{{\left( {{\lambda _d}} \right)}_P}} \over {{{\left( {{\lambda _d}} \right)}_e}}} = {{{m_e}} \over {5m}}$$ = 1.878 $$ \times $$ 10–4
$$ \Rightarrow $$ m = $${{{m_e}} \over {5 \times 1.874 \times {{10}^{ - 4}}}}$$
= $${{9.1 \times {{10}^{ - 31}}} \over {5 \times 1.874 \times {{10}^{ - 4}}}}$$
= 9.7 $$ \times $$ 10–28 kg
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