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  • (a) The highest energy level up to which the hydrogen atoms will be excited. (b) The longest wavelength in the (i) Lyman series, (ii) Balmer series of the spectrum of these hydrogen atoms.  

(a) The highest energy level up to which the hydrogen atoms will be excited.

(b) The longest wavelength in the (i) Lyman series, (ii) Balmer series of the spectrum of these hydrogen atoms.

 

Answers (1)

(a) Amount of energy required to excite the electron = 12.5eV

The energy of the electron in the nth state of an atom = \frac{-13.6z^{2}}{n^{2}}

For hydrogen. z=1

The energy required to excite an atom from the initial state (ni) to the final state (nf)

= \frac{-13.6z^{2}}{n^{2}_{f}}+\frac{-13.6z^{2}}{n^{2}_{i}}=12.5

n = 1 for the ground state of the hydrogen atom 

Energy of electron in ground state = -13.6 eV

\frac{-13.6z^{2}}{n^{2}_{f}}+13.6=12.5

1.1 = \frac{-13.6z^{2}}{n^{2}_{f}}

n_{f}^{2}=12.36

n_{f}=3.5

Since the state cannot be a fractional number

we have n_{f}=3

\therefore Hydrogen atom would be excited up to 3rd energy level.

(b) For Lyman series

\frac{1}{\lambda}= R \left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]

Rydberg constant, R = 1.097\times 10^{7}m^{-1}

For longest wavelength, n2=2, n1=1

\therefore \frac{1}{\lambda }=1.097\times 10^{7}\left [ \frac{1}{1^{2}}-\frac{1}{2^{2}} \right ]=8227500\Rightarrow \lambda =1215.4\AA

For Balmer series

\frac{1}{\lambda}= R \left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ] 

n1 = 2 , n1 = 1

\therefore \frac{1}{\lambda }=1.097\times 10^{7}\left [ \frac{1}{2^{2}}-\frac{1}{3^{2}} \right ]\Rightarrow \lambda = 6563\AA

Posted by

Safeer PP

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