Qstnid -Iv-3-a-hydrogen, (B) He+ And (C) Li++. (B) He+ And (C) Li++. - 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

with Solutions - page 3

Qstn# iv-3-a Prvs-Qstn Next-Qstn

#3-a
hydrogen, (b) He⁺ and (c) Li⁺⁺. (b) He⁺ and (c) Li⁺⁺.
Ans : Wavelength of the radiation emitted `` \left(\lambda \right)`` is given by
`` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
For hydrogen,
Atomic number, Z = 1
R = Rydberg constant = 1.097×10⁷ m^{`` -``1}
On substituting the respective values,
`` \lambda =\frac{1}{1.097\times {10}^{7}}=\frac{1}{1.097}\times {10}^{-7}``
`` =0.911\times {10}^{-7}``
`` =91.1\times {10}^{-9}=91\,\mathrm{\,nm\,}`` (b) For He⁺,
Atomic number, Z = 2
Wavelength of the radiation emitted by He⁺(`` \lambda ``) is given by
`` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \therefore \frac{1}{\lambda }={\left(2\right)}^{2}(1.097\times {10}^{7})\left(\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(\infty \right)}^{2}}\right)``
`` \Rightarrow \lambda =\frac{91\,\mathrm{\,nm\,}}{4}=23\,\mathrm{\,nm\,}`` (c) For Li⁺⁺,
Atomic number, Z = 3
Wavelength of the radiation emitted by Li⁺⁺(`` \lambda ``) is given by
`` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \therefore \frac{1}{\lambda }={\left(3\right)}^{2}\times (1.097\times {10}^{7})\left(\frac{1}{{1}^{2}}-\frac{1}{{\infty }^{2}}\right)``
`` \Rightarrow \lambda =\frac{91\,\mathrm{\,nm\,}}{{Z}^{2}}=\frac{91}{9}=10\,\mathrm{\,nm\,}``
Page No 384: (b) For He⁺,
Atomic number, Z = 2
Wavelength of the radiation emitted by He⁺(`` \lambda ``) is given by
`` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \therefore \frac{1}{\lambda }={\left(2\right)}^{2}(1.097\times {10}^{7})\left(\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(\infty \right)}^{2}}\right)``
`` \Rightarrow \lambda =\frac{91\,\mathrm{\,nm\,}}{4}=23\,\mathrm{\,nm\,}`` (c) For Li⁺⁺,
Atomic number, Z = 3
Wavelength of the radiation emitted by Li⁺⁺(`` \lambda ``) is given by
`` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \therefore \frac{1}{\lambda }={\left(3\right)}^{2}\times (1.097\times {10}^{7})\left(\frac{1}{{1}^{2}}-\frac{1}{{\infty }^{2}}\right)``
`` \Rightarrow \lambda =\frac{91\,\mathrm{\,nm\,}}{{Z}^{2}}=\frac{91}{9}=10\,\mathrm{\,nm\,}``
Page No 384:

#3-b
He⁺ and
Ans : For He⁺,
Atomic number, Z = 2
Wavelength of the radiation emitted by He⁺(`` \lambda ``) is given by
`` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \therefore \frac{1}{\lambda }={\left(2\right)}^{2}(1.097\times {10}^{7})\left(\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(\infty \right)}^{2}}\right)``
`` \Rightarrow \lambda =\frac{91\,\mathrm{\,nm\,}}{4}=23\,\mathrm{\,nm\,}``

#3-c
Li⁺⁺.
Ans : For Li⁺⁺,
Atomic number, Z = 3
Wavelength of the radiation emitted by Li⁺⁺(`` \lambda ``) is given by
`` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \therefore \frac{1}{\lambda }={\left(3\right)}^{2}\times (1.097\times {10}^{7})\left(\frac{1}{{1}^{2}}-\frac{1}{{\infty }^{2}}\right)``
`` \Rightarrow \lambda =\frac{91\,\mathrm{\,nm\,}}{{Z}^{2}}=\frac{91}{9}=10\,\mathrm{\,nm\,}``
Page No 384: