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The wavelength associated with the electron moving in the first orbit of hydrogen atom with velocity $2.2 \times 10^6 \mathrm{~ms}^{-1}$ (in nm) is
$\left(m_e=9.0 \times 10^{-31} \mathrm{~kg}, h=6.6 \times 10^{-34} \mathrm{Js}\right)$
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$\left(m_e=9.0 \times 10^{-31} \mathrm{~kg}, h=6.6 \times 10^{-34} \mathrm{Js}\right)$
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Verified Answer
The correct answer is:
$0.33$
According to de-Broglie's equation,
$\lambda=\frac{h}{m v}$
where, $\lambda$ is wavelength
$h$ is Planck's constant $\left(6.626 \times 10^{-34} \mathrm{Js}\right)$
$m$ is mass of electron (given $9.0 \times 10^{-31} \mathrm{~kg}$ )
$v$ is velocity of electron (given $2.2 \times 10^6 \mathrm{~m} / \mathrm{s}$ )
$\therefore \quad \lambda=\frac{6.626 \times 10^{-34}}{9.0 \times 10^{-31} \times 2.2 \times 10^6}$
$=0.33 \times 10^{-9} \mathrm{~m}$
i.e. $0.33 \mathrm{~nm}$ (as $1 \mathrm{~nm}=10^{-9} \mathrm{~m}$ )
$\lambda=\frac{h}{m v}$
where, $\lambda$ is wavelength
$h$ is Planck's constant $\left(6.626 \times 10^{-34} \mathrm{Js}\right)$
$m$ is mass of electron (given $9.0 \times 10^{-31} \mathrm{~kg}$ )
$v$ is velocity of electron (given $2.2 \times 10^6 \mathrm{~m} / \mathrm{s}$ )
$\therefore \quad \lambda=\frac{6.626 \times 10^{-34}}{9.0 \times 10^{-31} \times 2.2 \times 10^6}$
$=0.33 \times 10^{-9} \mathrm{~m}$
i.e. $0.33 \mathrm{~nm}$ (as $1 \mathrm{~nm}=10^{-9} \mathrm{~m}$ )
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