SATHEE: Structure of Atom - Result Question 15

Structure of Atom - Result Question 15

####15. The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of $H = 2.18 \times 10^{- 18} J^{- 1} a m^{- 1}$ and $h = 6.625 \times 10^{- 34} J s$ )

[2004]

(a) $1.54 \times 10^{15} s^{- 1}$

(b) $1.03 \times 10^{15} s^{- 1}$

(d) $2.00 \times 10^{15} s^{- 1}$

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Solution:

$2.18 \times 10^{- 18} = E_{1} - 0$

$E_{1} = 2.18 \times 10^{- 18} J^{- {atom}^{- 1}}$

$Δ E = E_{1} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

$\Rightarrow h v = 2.18 \times 10^{- 18} (\frac{1}{1^{2}} - \frac{1}{4^{2}})$

$\Rightarrow 6.625 \times 10^{- 34} \times v = 2.18 \times 10^{- 18} \times \frac{15}{16}$

$\Rightarrow v = \frac{2.18 \times 10^{- 18} \times 15}{6.625 \times 10^{- 34} \times 16}$

$\Rightarrow v = 3.08 \times 10^{15} s^{- 1}$

Structure of Atom - Result Question 15

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Structure of Atom - Result Question 15

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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