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A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of $3 \times 10^6 \mathrm{~ms}^{-1}$. The velocity of the particle is
(Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )
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(Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )
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Verified Answer
The correct answer is:
$2.7 \times 10^{-18} \mathrm{~ms}^{-1}$
Wavelength of a particle is given by
$\lambda=\frac{h}{p}$
where $h$ is Planck's constant.
and wavelength of an electron is given by
$\begin{aligned}
\lambda_e & =\frac{h}{p_e} \\
\lambda & =\lambda_e \\
p & =p_e \\
m v & =m_e y_e \\
y & =\frac{m_e v_e}{m}
\end{aligned}$
So, Putting the undergiven data, we get
$\begin{aligned}
& m_e=9.1 \times 10^{-31} \mathrm{~kg}, v_e=3 \times 10^6 \mathrm{~m} / \mathrm{s} \\
& m=1 \mathrm{mg}= 1 \times 10^{-6} \mathrm{~kg} \\
& y=\frac{9.1 \times 10^{-31} \times 3 \times 10^6}{1 \times 10^{-6}} \\
&=2.7 \times 10^{-18} \mathrm{~ms}^{-1}
\end{aligned}$
$\lambda=\frac{h}{p}$
where $h$ is Planck's constant.
and wavelength of an electron is given by
$\begin{aligned}
\lambda_e & =\frac{h}{p_e} \\
\lambda & =\lambda_e \\
p & =p_e \\
m v & =m_e y_e \\
y & =\frac{m_e v_e}{m}
\end{aligned}$
So, Putting the undergiven data, we get
$\begin{aligned}
& m_e=9.1 \times 10^{-31} \mathrm{~kg}, v_e=3 \times 10^6 \mathrm{~m} / \mathrm{s} \\
& m=1 \mathrm{mg}= 1 \times 10^{-6} \mathrm{~kg} \\
& y=\frac{9.1 \times 10^{-31} \times 3 \times 10^6}{1 \times 10^{-6}} \\
&=2.7 \times 10^{-18} \mathrm{~ms}^{-1}
\end{aligned}$
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