The frequency radiation emitted when the electron falls from n=4 to n=1 in a hydrogen atom will be (Given ionization energy of H=2.18 × 10^ -18 ~J atom^ -1 and h .=6.625 × 10^ -34 Js) :

Correct answer: C. 3.08 × 10^ 15 ~s^ -1 E=h v or v= E h For H atom, aligned E & = -21.76 × 10^ -19 n^2 ~J ~atm^ -1 & =-21.76 ×10^ -19 ((1)/(4^2)-(1)/(1^2)) aligned aligned & =20.40 ×10^ -19 ~J ~atm^ -1 v & = 20.40 × 10^ -19 6.625 × 10^ -34 & =3.079 ×10^ 15 ~s^ -1 . aligned

NEET 2004

Exam: NEET
Year: 2004
Subject: Chemistry
Chapter: Structure of Atom

NEET 2004 · Chemistry · Structure of Atom

The frequency radiation emitted when the electron falls from n=4 to...

The frequency radiation emitted when the electron falls from

\mathrm{n}=4

n=1

in a hydrogen atom will be (Given ionization energy of

\mathrm{H}=2.18 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}

and

h

\left.=6.625 \times 10^{-34} \mathrm{Js}\right)

Options

A $1.54 \times 10^{15} \mathrm{~s}^{-1}$
B $1.03 \times 10^{15} \mathrm{~s}^{-1}$
C $3.08 \times 10^{15} \mathrm{~s}^{-1}$
D $2.00 \times 10^{15} \mathrm{~s}^{-1}$

Correct answer

C. $3.08 \times 10^{15} \mathrm{~s}^{-1}$

Step-by-step solution

\mathrm{E}=h \mathrm{v}

or

\mathrm{v}=\frac{\mathrm{E}}{h}

For

\mathrm{H}

atom,

\begin{aligned}E & =\frac{-21.76 \times 10^{-19}}{n^2} \mathrm{~J} \mathrm{~atm}^{-1}\\ \DeltaE & =-21.76 \times10^{-19}\left(\frac{1}{4^2}-\frac{1}{1^2}\right) \end{aligned}

$\begin{aligned}& =20.40 \times10^{-19} \mathrm{~J} \mathrm{~atm}^{-1}\\ v & =\frac{20.40 \times 10^{-19}}{6.625 \times 10^{-34}}\\ & =3.079 \times10^{15} \mathrm{~s}^{-1} . \end{aligned}

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