Question

1. a) Calculate the ionisation energy of rubidium per atom, if light of wavelength 5.84 x 10 raised to the power of -8 m produces electrons with a speed of 2.450 x 10 raised to the power of 6 m/sec.
[Hint: Assume that the threshold frequency refers to the frequency corresponding to the ionisation energy.]


b) Assume that the electron in Li raised to the power 2+ ion is in third orbit. Calculate
i) the radius of the orbit and,
ii) the total energy of the electron.

[Hint: Li raised to the power 2+ ion also has atomic spectra similar to hydrogen atom. While applying relevant equations, use Z = 3.]

Answer 1

Answer on Question #62022-Physics-Atomic and Nuclear Physics

1. a) Calculate the ionization energy of rubidium per atom, if light of wavelength $5.84 \times 10$ raised to the power of $-8 \mathrm{~m}$ produces electrons with a speed of $2.450 \times 10$ raised to the power of $6 \mathrm{~m} / \mathrm{sec}$.

[Hint: Assume that the threshold frequency refers to the frequency corresponding to the ionization energy.]

Solution

$$E _ {l i g h t} = E _ {i o n i s a t i o n} + E _ {k i n e t i c}$$ $$E _ {i o n i s a t i o n} = E _ {l i g h t} - E _ {k i n e t i c}$$ $$E _ {i o n i s a t i o n} = \frac {h c}{\lambda} - \frac {m v ^ {2}}{2} = \frac {6 . 6 2 6 \cdot 1 0 ^ {- 3 4} \cdot 3 \cdot 1 0 ^ {8}}{5 . 8 4 \cdot 1 0 ^ {- 8}} - \frac {9 . 1 1 \cdot 1 0 ^ {- 3 1} (2 . 4 5 0 \cdot 1 0 ^ {6}) ^ {2}}{2} = 6. 6 7 \cdot 1 0 ^ {- 1 9} J.$$

b) Assume that the electron in Li raised to the power $2+$ ion is in third orbit. Calculate

i) the radius of the orbit and,

ii) the total energy of the electron.

[Hint: Li raised to the power $2+$ ion also has atomic spectra similar to hydrogen atom. While applying relevant equations, use $Z = 3$.]

Solution

(i) the radius of the orbit

$$r = \frac {h ^ {2} n ^ {2}}{4 \pi m k Z e ^ {2}} = \frac {(6 . 6 \cdot 1 0 ^ {- 3 4}) ^ {2} \cdot (3) ^ {2}}{4 \pi \cdot 9 . 1 \cdot 1 0 ^ {- 3 1} \cdot 9 \cdot 1 0 ^ {9} \cdot 3 (1 . 6 \cdot 1 0 ^ {- 1 9}) ^ {2}} = 5. 0 \cdot 1 0 ^ {- 1 0} m.$$

(ii) the total energy of the electron

$$E = - \frac {1}{2} \frac {k Z e ^ {2}}{r} = - \frac {1}{2} \frac {9 \cdot 1 0 ^ {9} \cdot 3 (1 . 6 \cdot 1 0 ^ {- 1 9}) ^ {2}}{5 . 0 \cdot 1 0 ^ {- 1 0}} = - 6. 9 \cdot 1 0 ^ {- 1 9} J$$

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