A Hydrogen atom in its ground state is irradiated by light of wavelength 970 A0 . Taking hc/e=1.237×10−6 eV m and the ground state energy of hydrogen atom as -13.6 eV, the number of lines present in the emission spectrum is

E= hcπ=12370970=12.7 eVThe incident light excite electron to n quantum state ⇒ −13.6+12.7=−13.6n2n2=16 or n=4 Number of lines =nC2=n(n−1)2=6

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A Hydrogen atom in its ground state is irradiated by light of wavelength 970 $\overset{0}{A}$ . Taking $h c / e = 1.237 \times 10^{- 6} e V m$ and the ground state energy of hydrogen atom as -13.6 eV, the number of lines present in the emission spectrum is

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Detailed Solution

$E = \frac{h c}{π} = \frac{12370}{970} = 12.7 e V$

The incident light excite electron to n quantum state $\Rightarrow - 13.6 + 12.7 = - \frac{13.6}{n^{2}}$

$n^{2} = 16 o r n = 4$

Number of lines $=^{n} C_{2} = \frac{n (n - 1)}{2} = 6$

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