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Question: The frequency of radiation emitted when an electron falls from n = 4 to n = 1 in a H - atom will be?...

The frequency of radiation emitted when an electron falls from n = 4 to n = 1 in a H - atom will be? (Given that: ionisation energy of H atom : 2.18×10−18Jatom−12.18 \times {10^{ - 18}}Jato{m^{ - 1}}and Planck's constant (h) = 6.625×10−34Js6.625 \times {10^{ - 34}}Js
A) 1.03×103s11.03 \times {10^3}{s^1}
B) 3.084×1015s13.084 \times {10^{15}}{s^1}
C) 2×1015s12 \times {10^{15}}{s^1}
D) – 3.435 J

Explanation

Solution

The atomic number of H – atom is 1 which will give Z = 1, so we can give the energy in orbit by the division of that in first orbit with square of 1 (as orbit number n = 1). We can calculate the change in energy of falling the atom from one orbit to another and then apply Planck’s equation to calculate the required value of frequency.

Formula Used: Planck’s equation:
ΔE=hν\Delta E = h\nu where ΔE\Delta E is the change in energy, h is thee planck’s constant and ν\nu is the frequency.

Complete step by step answer: Ionization energy means it is the energy for the first orbit is given to be 2.18×10−18Jatom−12.18 \times {10^{ - 18}}Jato{m^{ - 1}}. Atomic number of hydrogen atom is 1 i.e. Z = 1, then energy for nth{n^{th}} orbit is given as:
En=−2.18×10−18n2{E_n} = \dfrac{{ - 2.18 \times {{10}^{ - 18}}}}{{{n^2}}} _________ (1)
For n = 1
E=−2.18×10−18E = - 2.18 \times {10^{ - 18}}
The change in energy when electrons fall from n = 4 to n = 1 is given as:
ΔE=E4−E1\Delta E = {E_4} - {E_1}
Then, using (1):
ΔE=−2.18×10−1842−(−2.18×10−1812) ⇒ΔE=−2.18×10−18(142−112) ⇒ΔE=−2.18×10−18×(−1516) ⇒ΔE=2.0437×10−18Jatom−1  \Delta E = \dfrac{{ - 2.18 \times {{10}^{ - 18}}}}{{{4^2}}} - \left( {\dfrac{{ - 2.18 \times {{10}^{ - 18}}}}{{{1^2}}}} \right) \\\ \Rightarrow \Delta E = - 2.18 \times {10^{ - 18}}\left( {\dfrac{1}{{{4^2}}} - \dfrac{1}{{{1^2}}}} \right) \\\ \Rightarrow \Delta E = - 2.18 \times {10^{ - 18}} \times \left( { - \dfrac{{15}}{{16}}} \right) \\\ \Rightarrow \Delta E = 2.0437 \times {10^{ - 18}}Jato{m^{ - 1}} \\\
According to Planck’s equation the change in energy is directly proportional to frequency of electron and Planck’s constant (h) is used as the proportionality constant. So:
ΔE=hν ⇒ν=ΔEh  \Delta E = h\nu \\\ \Rightarrow \nu = \dfrac{{\Delta E}}{h} \\\
Substituting the known values to find the value of frequency, we get:
ν=2.0437×10−186.625×10−34 ⇒ν=3.084×1015s−1  \nu = \dfrac{{2.0437 \times {{10}^{ - 18}}}}{{6.625 \times {{10}^{ - 34}}}} \\\ \Rightarrow \nu = 3.084 \times {10^{15}}{s^{ - 1}} \\\
Therefore, the frequency of radiation emitted when an electron falls from n = 4 to n = 1 in a H atom will be 3.084×1015s13.084 \times {10^{15}}{s^1} and the correct option is B).

Note: Joules (J) is the SI unit of energy, but frequency is basically measured per second, thus we can eliminate option D) just by seeing its units. Units play an important role in identifying the quantities.