In the ground state of hydrogen atom. electron absorbs 1.5 times energy than the minimum energy $\left(2.18…
In the ground state of hydrogen atom. electron absorbs 1.5 times energy than the minimum energy $\left(2.18 \times 10^{-18} \mathrm{~J}\right)$ to escape from the atom. The wavelength of the emitted electron (in m) is $\left(m_e=9 \times 10^{-31} \mathrm{~kg}\right)$
$\frac{h \times 10^{24}}{\sqrt{1.962}}$
$\frac{h}{\sqrt{1.962}} \times 10^{23}$
$\frac{h}{\sqrt{1.962}} \times 10^{25}$
$\frac{h}{\sqrt{1.962}} \times 10^{22}$
Solution
Energy absorbed by the element
$\begin{aligned}
& =1.5 \times 2.18 \times 10^{-18} \mathrm{~J} \\
& =1.5 \times 2.18 \times 10^{-18} \mathrm{kgm}^2 \mathrm{~s} \mathrm{~s}^{-2} \\
& =3.27 \times 10^{-18} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}
\end{aligned}$
Energy required $=2.18 \times 10^{-18} \mathrm{~J}$
Kinetic energy of the emitted electron
$\begin{aligned}
& =\text { Energy absorbed }- \text { energy required } \\
& =\left(3.27 \times 10^{-18}-2.18 \times 10^{-18}\right) \mathrm{kg} \mathrm{~m}^2 \mathrm{~s}^{-2} \\
& \mathrm{KE}=1.09 \times 10^{-18} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2} \\
& \quad \mathrm{~m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}
\end{aligned}$
According to de Broglie equation
$\lambda=\frac{\mathrm{h}}{\mathrm{~m} v} \mathrm{KE}=\frac{1}{2} \mathrm{~m} v^2$
$\begin{aligned}
& v=\left(\frac{2 \times \mathrm{KE}}{\mathrm{~m}}\right)^{1 / 2} \\
& \therefore \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK} \cdot \mathrm{E}}}
\end{aligned}$
or, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \times 9 \times 10^{-31} \mathrm{~kg} \times 1.09 \times 10^{-18} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}}}$
or, $\lambda=\frac{h}{\sqrt{19.62 \times 10^{-49}}}$
$\lambda=\frac{\mathrm{h}}{\sqrt{1.962 \times 10^{-48}}} \Rightarrow \lambda=\frac{\mathrm{h} \times 10^{24}}{\sqrt{1.962}}$