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Wave Optics

Question

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The ground state energy of hydrogen atoms is – 13.6 eV.
(i) Which are the potential and kinetic energy of an electron in the third excited state?
(ii) If the electron jumps to the ground state from the third excited state. Calculate the frequency of photon emitted.

Solution

Given, ground state of hydrogen atom = -13.6 eV

(i) Ground state energy in hydrogen atom is given by,

E = - (K . E .) = \frac{1}{2} (P . E .)

∴ Potential energy of electron in third excited state,

= \frac{2 (E)}{(4)^{2}} = - \frac{2 \times 13.6}{16} eV = - 1.7 eV

and,

Kinetic energy of an electron,

= - \frac{1}{2} (1.7) = 0.85 eV

(ii) When the electron jumps from the ground state to the third excited state,

Frequency of the photon ,

v = \frac{∆ E}{h}

= [\frac{- 13.6}{(4)^{2}} + \frac{13.6}{(1)^{2}}] \frac{1.6 \times 10^{- 19}}{6.6 \times 10^{- 34}}

=

\frac{11.9 \times 1.6 \times 10^{- 19}}{6.6 \times 10^{- 34}} Hz \approx 2.9 \times 10^{15} Hz .

Some More Questions From Wave Optics Chapter

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Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson's model much less, about the same, or much greater than that predicted by Rutherford's model?

Answer the following questions, which help you understand the difference between Thomson's model and Rutherford's model better.
Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

Answer the following questions, which help you understand the difference between Thomson's model and Rutherford's model better.
In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil?

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