The ionization energy of hydrogen atom is 13.6eV . Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy 12.1eV . How many spectral lines will be emitted by the hydrogen atoms

Question

The ionization energy of hydrogen atom is 13.6eV . Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy 12.1eV . How many spectral lines will be emitted by the hydrogen atoms (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Energy required for transition of electron is given by,

Δ E = E_{0} - E_{n}

\Rightarrow 12.1 = 13.6 - E_{n}

\Rightarrow E_{n} = 13.6 - 12.1 = 1.5

\Rightarrow \frac{13.6}{n ^{2}} = 1.5

\Rightarrow n^{2} = \frac{13.6}{1.5} = 9

\Rightarrow n = 3

Number of spectral lines

= \frac{n ( n - 1 )}{2} = \frac{3 ( 3 - 1 )}{2} = 3

Q. The ionization energy of hydrogen atom is $13.6 e V$ . Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy $12.1 e V$ . How many spectral lines will be emitted by the hydrogen atoms

$1$

$2$

$3$

$4$

Q. The ionization energy of hydrogen atom is 13.6eV . Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy 12.1eV . How many spectral lines will be emitted by the hydrogen atoms

1

2

3

4

Energy required for transition of electron is given by, ΔE=E0​−En​ ⇒12.1=13.6−En​ ⇒En​=13.6−12.1=1.5 ⇒n213.6​=1.5 ⇒n2=1.513.6​=9 ⇒n=3 Number of spectral lines =2n(n−1)​=23(3−1)​=3

Q. The ionization energy of hydrogen atom is $13.6 e V$ . Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy $12.1 e V$ . How many spectral lines will be emitted by the hydrogen atoms

$1$

$2$

$3$

$4$