The velocity of the electron in Bohr's first orbit is $x \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$. The…
The velocity of the electron in Bohr's first orbit is $x \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$. The de-Broglie wavelength associated with it (in nm) is $\left(m_e=9 \times 10^{-31} \mathrm{~kg}, h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)$
$\frac{x}{1.43}$
$\frac{x}{0.73}$
$\frac{0.73}{x}$
$\frac{0.073}{x}$
Solution
According to de-Broglie's equation,
$\lambda=\frac{h}{m_e v}$
$\begin{aligned} & \lambda=\text { de-Broglie wavelength } \\ & h=\text { Planck's constant }=6.626 \times 10^{-34} \mathrm{JS}\end{aligned}$
$\begin{aligned} m_e & =\text { mass of electron }=9.1 \times 10^{-31} \mathrm{~kg} \\ v & =\text { velocity of electron }=x \times 10^6 \mathrm{~m} / \mathrm{s}\end{aligned}$
$\begin{aligned} & =\frac{6.626 \times 10^{-34} \mathrm{Js}}{9.1 \times 10^{-31} \mathrm{~kg} \times x \times 10^6 \mathrm{~m} / \mathrm{s}} \\ & =\frac{0.728 \times 10^{-9} \mathrm{~m}}{x}\end{aligned}$
or $\lambda=\frac{0.728}{x} \mathrm{~nm}$