Modern Physics 1 Question 52
48. An electron in a hydrogen like atom is in an excited state. It has a total energy of $-3.4 eV$. Calculate
(1996, 3M)
(a) the kinetic energy,
(b) the de-Broglie wavelength of the electron.
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Answer:
Correct Answer: 48. (a) $3.4 eV$ (b) $6.63 \AA \quad \mathbf{4 9 . 6 , 3}$
Solution:
- (a) Kinetic energy of electron in the orbits of hydrogen and hydrogen like atoms $=\mid$ Total energy $\mid$
$$ \therefore \quad \text { Kinetic energy }=3.4 eV $$
(b) The de-Broglie wavelength is given by
$$ \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 K m}} $$
Here, $K=$ kinetic energy of electron Substituting the values, we have
$$ \begin{aligned} & \lambda=\frac{\left(6.6 \times 10^{-34} J-s\right)}{\sqrt{2\left(3.4 \times 1.6 \times 10^{-19} J\right)\left(9.1 \times 10^{-31} kg\right)}} \\ & \lambda=6.63 \times 10^{-10} m \\ & \text { or } \quad \lambda=6.63 \AA \end{aligned} $$
