SATHEE: atoms Question 19

atoms Question 19

Question: Q. 10. Show that the radius of the orbit in hydrogen atom varies as $n^{2}$ , where $n$ is the principal quantum number of the atom.

U] [Delhi I, II, III 2015]

$\begin{aligned} (i) & \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{r^{2}} & = \frac{m v^{2}}{r} r & = \frac{e^{2}}{4 π ε_{0} m v^{2}} m v^{2} r & = \frac{e^{2}}{4 π ε_{0}} \end{aligned}$

According to the Bohr’s postulate,

$\begin{aligned} m v r & = \frac{n h}{2 π} m^{2} v^{2} r^{2} & = \frac{n^{2} h^{2}}{4 π^{2}} \end{aligned}$

Putting the value of $m v^{2} r$ from eqn. (i)

$\begin{aligned} \frac{e^{2}}{4 π ε_{0}} m r & = \frac{n^{2} h^{2}}{4 π^{2}} r & = (\frac{n^{2}}{m}) (\frac{h}{2 π}) \frac{4 π ε_{0}}{e^{2}} \end{aligned}$

The above equation shows that $r$ is directly proportional to $n^{2}$

[CBSE Marking Scheme 2015]

Commonly Made Error

Some students were unable to recall the correct equation.

i.e., $m v r = \frac{n h}{2 π}$ and $\frac{m v^{2}}{r} = \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{r^{2}}$

Q. 11. The electron, in a hydrogen atom, is in its second excited state.

Calculate the wavelength of the lines in the Lyman series, that can be emitted through the permissible transitions of this electron.

(Given the value of Rydberg constant, $R = 1.1 \times$ $10^{7} m^{- 1}$ )

A [Delhi Comptt., I, II, III 2016]

Show Answer

Solution:

Ans. For second excited state, $n = 3$

Hence two possible transition of the Lyman series : $3 \to 1$ and $2 \to 1$ .

Wavelength for transition $3 \to 1, n_{f} = 1, n_{i} = 3$

$\begin{aligned} \frac{1}{λ} & = R (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}) \frac{1}{λ} & = 1.1 \times 10^{7} (\frac{1}{1} - \frac{1}{9}) & = 1.1 \times 10^{7} (\frac{8}{9}) \Rightarrow λ & = \frac{9}{8 \times 1.1 \times 10^{- 7}} & = 1.023 \times 10^{- 7} & = 102.3 nm \end{aligned}$

For transition $2 \to 1, n_{f} = 1, n_{i} = 2$

$\frac{1}{λ} = 1.1 \times 10^{7} (1 - \frac{1}{4})$

$\Rightarrow$

$λ = 212 nm$

[CBSE Marking Scheme 2016]

Commonly Made Error

Many students couldn’t remember that in Lyman series. $n_{f} = 1$

Answering Tips

Students should learn either with energy level diagram, or by drawing 7-8 orbits that how various series of lines are formed in the hydrogen spectrum.

Atoms

atoms Question 19

Commonly Made Error

Commonly Made Error

Answering Tips

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Atoms

atoms Question 19

Commonly Made Error

Commonly Made Error

Answering Tips

Table of Contents

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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