Qstnid -Iv-20-suppose, In Certain Conditions Only Those Transitions Are Allowed To Hydrogen Atoms In Which The Principal Quantum Number N Changes By 2. (A) Find The Smallest Wavelength Emitted By Hydrogen. (B) List The Wavelength Emitted By Hydrogen In The Visible Range (380 Nm To 780 Nm). (A) Find The Smallest Wavelength Emitted By Hydrogen. (B) List The Wavelength Emitted By Hydrogen In The Visible Range (380 Nm To 780 Nm). - 43: Bohr's Model And Physics Of The Atom

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NEET-XII-Physics

43: Bohr's Model and Physics of the Atom

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#20
Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm). (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).
Ans : Given:
Possible transitions:
From n₁ = 1 to n₂ = 3
n₁ = 2 to n₂ = 4 (a) Here, n₁ = 1 and n₂ = 3
Energy, `` E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` E=13.6\left(\frac{1}{1}-\frac{1}{9}\right)``
`` =13.6\times \frac{8}{9}.....\left(1\right)``
`` \,\mathrm{\,Energy\,}\left(E\right)\,\mathrm{\,is\,}\,\mathrm{\,also\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,}``
`` E=\frac{hc}{\,\mathrm{\,\lambda \,}}``
`` ``
Here, h = Planck constant
c = Speed of the light
`` \lambda `` = Wavelength of the radiation
`` ``
`` \therefore E=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{\lambda }.....\left(2\right)``
`` \,\mathrm{\,Equating\,}\,\mathrm{\,equation\,}s\left(1\right)\,\mathrm{\,and\,}\left(2\right),\,\mathrm{\,we\,}\,\mathrm{\,have\,}``
`` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}\times 9}{13.6\times 8}``
`` =0.027\times {10}^{-7}=103\,\mathrm{\,nm\,}`` (b) Visible radiation comes in Balmer series.
As 'n' changes by 2, we consider n = 2 to n = 4.
Energy, `` {E}_{1}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` =13.6\times \left(\frac{1}{4}-\frac{1}{16}\right)``
`` =2.55\,\mathrm{\,eV\,}``
If `` {\lambda }_{1}`` is the wavelength of the radiation, when transition takes place between quantum number n = 2 to n = 4, then
255 = `` \frac{1242}{{\lambda }_{1}}``
or λ₁ = 487 nm
Page No 384: (a) Here, n₁ = 1 and n₂ = 3
Energy, `` E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` E=13.6\left(\frac{1}{1}-\frac{1}{9}\right)``
`` =13.6\times \frac{8}{9}.....\left(1\right)``
`` \,\mathrm{\,Energy\,}\left(E\right)\,\mathrm{\,is\,}\,\mathrm{\,also\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,}``
`` E=\frac{hc}{\,\mathrm{\,\lambda \,}}``
`` ``
Here, h = Planck constant
c = Speed of the light
`` \lambda `` = Wavelength of the radiation
`` ``
`` \therefore E=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{\lambda }.....\left(2\right)``
`` \,\mathrm{\,Equating\,}\,\mathrm{\,equation\,}s\left(1\right)\,\mathrm{\,and\,}\left(2\right),\,\mathrm{\,we\,}\,\mathrm{\,have\,}``
`` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}\times 9}{13.6\times 8}``
`` =0.027\times {10}^{-7}=103\,\mathrm{\,nm\,}`` (b) Visible radiation comes in Balmer series.
As 'n' changes by 2, we consider n = 2 to n = 4.
Energy, `` {E}_{1}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` =13.6\times \left(\frac{1}{4}-\frac{1}{16}\right)``
`` =2.55\,\mathrm{\,eV\,}``
If `` {\lambda }_{1}`` is the wavelength of the radiation, when transition takes place between quantum number n = 2 to n = 4, then
255 = `` \frac{1242}{{\lambda }_{1}}``
or λ₁ = 487 nm
Page No 384:

#20-a
Find the smallest wavelength emitted by hydrogen.
Ans : Here, n₁ = 1 and n₂ = 3
Energy, `` E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` E=13.6\left(\frac{1}{1}-\frac{1}{9}\right)``
`` =13.6\times \frac{8}{9}.....\left(1\right)``
`` \,\mathrm{\,Energy\,}\left(E\right)\,\mathrm{\,is\,}\,\mathrm{\,also\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,}``
`` E=\frac{hc}{\,\mathrm{\,\lambda \,}}``
`` ``
Here, h = Planck constant
c = Speed of the light
`` \lambda `` = Wavelength of the radiation
`` ``
`` \therefore E=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{\lambda }.....\left(2\right)``
`` \,\mathrm{\,Equating\,}\,\mathrm{\,equation\,}s\left(1\right)\,\mathrm{\,and\,}\left(2\right),\,\mathrm{\,we\,}\,\mathrm{\,have\,}``
`` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}\times 9}{13.6\times 8}``
`` =0.027\times {10}^{-7}=103\,\mathrm{\,nm\,}``

#20-b
List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).
Ans : Visible radiation comes in Balmer series.
As 'n' changes by 2, we consider n = 2 to n = 4.
Energy, `` {E}_{1}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` =13.6\times \left(\frac{1}{4}-\frac{1}{16}\right)``
`` =2.55\,\mathrm{\,eV\,}``
If `` {\lambda }_{1}`` is the wavelength of the radiation, when transition takes place between quantum number n = 2 to n = 4, then
255 = `` \frac{1242}{{\lambda }_{1}}``
or λ₁ = 487 nm
Page No 384: