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Question: Calculate the wavelength in angstrom of the photon that is emitted when an electron in Bohr orbit n=...

Calculate the wavelength in angstrom of the photon that is emitted when an electron in Bohr orbit n=2 returns to the orbit n=1 in the hydrogen atom. The ionization potential of the ground state of the hydrogen atom is 2.17×10−11ergatom−12.17\times {{10}^{-11}}ergato{{m}^{-1}}.

Explanation

Solution

Hint : Bohr produces a set of postulates to explain the hydrogen atom. We know the energy difference between two energy levels is equal to the energy of a photon and its equation is given as ΔE=hν\Delta E=h\nu , where ΔE\Delta E is the energy difference between two energy levels, h is the Planck’s constant and ν\nu is the frequency.

Complete step by step solution :
In the question, we have to find the difference between two energy levels, which give the wavelength of the photon that is emitted. This is explained by using a formula called Rydberg formula.
E=R(1nf2−1ni2)E=R(\dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}) Here, nfn_{f}^{{}} is the final energy level, that isnf=2n_{f}^{{}}=2. And nin_{i}^{{}}is the initial energy level, that is ni=1n_{i}^{{}}=1 and E is the energy level and R is the Rydberg constant. So, For 2 energy levelsE2{{E}_{2}} and E2{{E}_{2}} .
E2−E1=−2.17×10−11(14−11){{E}_{2}}-{{E}_{1}}=-2.17\times {{10}^{-11}}(\dfrac{1}{4}-\dfrac{1}{1})=2.17×10−11342.17\times {{10}^{-11}}\dfrac{3}{4}
Here, nfn_{f}^{{}}is the final energy level, that is nf=2n_{f}^{{}}=2. And nin_{i}^{{}}is the initial energy level, that is ni=1n_{i}^{{}}=1
E2−E1=ΔE=hv=cλh{{E}_{2}}-{{E}_{1}} = \Delta E=hv=\dfrac{c}{\lambda }h
We can write the above equation in another manner too based on wavelength.
λ=hcΔE\lambda = \dfrac{hc}{\Delta E}
Where, λ\lambda is the wavelength, h is the Planck’s constant, c is the velocity of light and ΔE\Delta E difference in energy levels. We know, h=6.62×10−27ergh=6.62\times {{10}^{-27}}erg, c=3×1010cmc=3\times {{10}^{10}}cm and ΔE\Delta E is given as2.17×10−11342.17\times {{10}^{-11}}\dfrac{3}{4}.
λ=6.62×10−27erg×3×1010cm2.17×10−1134=2.6482.1710−5cm\lambda =\dfrac{6.62\times {{10}^{-27}}erg\times 3\times {{10}^{10}}cm}{2.17\times {{10}^{-11}}\dfrac{3}{4}} = \dfrac{2.648}{2.17}{{10}^{-5}}cm
=1.22×10−5cm=1220A˚=1.22\times { 10 }^{ -5 }cm = 1220\mathring { A }

Additional Information:
Niels Bohr produces the atomic Hydrogen model. In it he describes, a positively charged nucleus consists of protons and neutrons and is surrounded by negatively charged electron clouds. The atom is held together by electrostatic forces between the positively charged nucleus and negatively charged surrounding, that is electrons. The structure of hydrogen in Bohr’s model has energy levels.

Note : In the question, wavelength in angstrom is asked. We are calculating wavelength in centimetres. We should always see to it that it is converted to angstrom units. 1 Armstrong =10−8{{10}^{-8}} centimetres.