Question

If the wavelength for an electron emitted from H-atom is 3.3 × 10^–10 m, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is ______ times. (Nearest integer)

\(\begin{array}{l}\left[\text{Given} : h = 6.626 \times 10^{-34}~ \text{J s}\right]\\ \left[\text{Mass of electron} = 9.1 \times 10^{-31} ~\text{kg}\right]\end{array}\)

Answer 1

We know that,\( \begin{array}{l} \lambda =\frac{h}{mv} \end{array}\)

\(\begin{array}{l} \Rightarrow mv=\frac{h}{\lambda}=\frac{6.626\times10^{-34}\text{kg}\frac{m^2}{\sec^2}\times\sec}{3.3\times10^{-10}\text{ m}}\end{array}\)

\(\begin{array}{l} mv=\frac{6.626\times10^{-24}}{3.3}=2\times 10^{-24}~\text{kg m}\sec^{-1}\end{array}\)

\(\begin{array}{l} \text{Kinetic energy}=\frac{1}{2}\text{ mv}^2 \end{array}\)

\(\begin{array}{l} =\frac{\left(mv\right)^2}{2m} \end{array}\)\( \begin{array}{l} =\frac{\left(2\times10^{-24}\right)^2}{2\times9.1\times10^{-31}\text{kg}}\end{array}\)

\(\begin{array}{l}= 2.18 \times 10^{-18}\text{J}\\ = 21.8 \times 10^{-19} \text{J}\end{array}\)

\(\begin{array}{l} \begin{matrix}\text{Total energy} & = & \text{lonization} & + & \text{Kinetic} \\\text{absorbed} & & \text{energy} & &\text{energy} \\\end{matrix}\end{array}\)

\(\begin{array}{l}= \left(21.76 + 21.8\right) \times 10^{-19}\\ = 43.56 \times 10^{-19}~ \text{J}\\ \approx 2 ~\text{times of}~ 21.76 \times 10^{-19} ~\text{J} \end{array}\)