Atoms Test

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Atoms Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

In an atom an electron excites to the fourth orbit. When it jumps back to the energy levels a spectrum is formed.Total number of spectral lines in this spectrum would be

A
3

B
4

C
5

D
6

Solution

In an atom an electron excites to the fourth orbit, $$n=4$$
The total number off spectral lines in the spectrum is,
$$\dfrac{n(n-1)}{2}=\dfrac{4(4-1)}{2}=\dfrac{4\times 3}{2}=6$$
The correct option is D.
Question 2
1 / -0

A proton, a direction ion and an $$\alpha$$-particle of equal kinetic energy perform circular motion normal to a uniform magnetic field $$B$$. If the radii of their paths are $$r_p, r_d$$ and $$r_\alpha$$ respectively then [Here, $$q_d = q_p, m_d = 2m_p$$]

A
$$r_\alpha = r_p > r_d$$

B
$$r_\alpha = r_p< r_d$$

C
$$r_\alpha > r_d > r_p$$

D
$$r_\alpha = r_d = r_p$$

Solution

$$\begin{array}{l}\text { Given, } \\\text { a proton } \quad q=+1 \quad m=m_p \\\text { a deutorium } \quad q=+ 1 \quad m=2 \mathrm{m_p} \\\text { a } \alpha \text { particle }\quad q=+ 2 \quad m=4 \mathrm{m_p} \end{array}$$
$$\begin{array}{l}\text { hane equal Kinctic energy } \\\qquad B=\text { magnetic field. }\end{array}$$
$$\begin{array}{l}\text { We Know when a change particle enters } \\\text { in a magnetic field it perform } \\\text { circular motion with radius equal to. }\end{array}$$
$$\begin{aligned}r &=\frac{m v}{q B} \\r &=\frac{\sqrt{2 m K E}}{q B}\end{aligned}$$
$$\begin{array}{l}r_{\text {proton }}=\frac{\sqrt{2 \times m_p \times K E}}{1\times B} \\\text r_{ deutorium }=\frac{\sqrt{2 \times 2 \times m_p \times K E}}{1 \times B}\\\text r_{\alpha }=\frac{\sqrt{2 \times 4 \times m_p\times K E}}{2 \times B}=\frac{\sqrt{2 \times m_p \times K E}}{1 \times B}\end{array}$$
$$\text { So, } \quad r_{\text {proton }}=r_{\alpha} < r_{\text {deutorium }}(\text {option } B)$$
Question 3
1 / -0

If radiation of all wavelengths from ultraviolet to infrared is passed through hydrogen gas at room temperature absorption lines will be observed in the

A
Lyman series

B
Balmer's series

C
Both (1) and (2)

D
Neither (1) or (2)

Solution

$$\begin{array}{l}\text { For hydrogen atom only one electron is } \\\text { present at ground state. when it absorb } \\\text { energy of different wavelength, it transist } \\\text { from } n=1 \text { to } n=2,3,4 \text { and so on, } \\ \text{so, only lymen series id obsererved due to}\\\text { absence of electron on other energy level. }\end{array}$$
Question 4
1 / -0

The radius of first orbit of hydrogen atom is $$0.53\ A$$. The radius of its fourth orbit will be-

A
$$0.193\ A$$

B
$$4.24\ A$$

C
$$2.12\ A$$

D
$$8.48\ A$$

Solution

$$\text{Given- $*$ Radius of first orbit=0.53 A}$$
$$\text{We know that -}$$
$$\text{Radius of the orbit of $e^{-}(r),$}$$
$$\text { r } \alpha n^{2} ,\quad n \rightarrow n^{\text {th }}orbit$$
$$\Rightarrow \frac{r_{1}}{r_{2}}=\frac{n_{1}^{2}}{n_{2}^{2}}\quad\text { Let } r_{2} \text { be the radius of } 4^{\text {th }} \text { orbit }$$
$$\Rightarrow \quad r_{2}=\frac{n_{2}^{2}}{n_{1}^{2}} \cdot r_{1}$$

$$\begin{array}{l}\text { for } 4^{\text {th }} \text { orbit, }n_{2}=4\text { and } n_{1}=1 \\\Rightarrow \quad r_{2}=(4)^{2} \cdot r_{1}=16 \times 0.53 A^{\circ} \\\Rightarrow r_{2}=8.48 A^{\circ}\\\text { Hence, option (D) is correct. }\end{array}$$
Question 5
1 / -0

If in Bohr's atomic model, it is assumed that force between electron the proton varies inversely as $${r^4},$$b energy of the system will be proportional to:

A
$${n^2}$$

B
$${n^4}$$

C
$${n^6}$$

D
$${n^8}$$

Solution

$$\dfrac{mv^2}{r}=\dfrac{k}{r^4}$$
$$V^2=\dfrac{k}{mr^3}$$
we know
$$mvr=\dfrac {nh}{2x}$$
$$m^2v^2r^2=\dfrac{n^2h^2}{4\pi^2}$$
$$\dfrac{m^2kr^2}{mr^3}=\dfrac{n^2h^2}{4x^2}$$
$$r=\dfrac{mk4\pi^2}{n^2h^2}$$
$$k.f$$=$$\dfrac{1}{2}mv^2=\dfrac{1}{2}m/\dfrac{k}{mr^3}$$
$$=\dfrac{1}{2}m\dfrac{K}{m^1m^3}$$
$$E\alpha\ n^6$$
Question 6
1 / -0

If m is mass of electron, v its velocity, r the radius of stationary circular orbit around a nucleus with charge Ze then from Bohr's first postulate, the kinetic energy $$K=\dfrac{1}{2}mv^2$$ of the electron in C.G.S. system is equal to?

A
$$\dfrac{1}{2} \dfrac{Ze^2}{r}$$

B
$$\dfrac{1}{2}\dfrac{Ze^2}{r^2}$$

C
$$\dfrac{Ze^2}{r}$$

D
$$\dfrac{Ze^2}{r^2}$$

Solution

In the revolution of an electron, coulomb force provides the necessary centripetal force

$$\Rightarrow \dfrac{ze^{2}}{r^{2}}=\dfrac{mv^{2}}{r}\Rightarrow mv^{2}=\dfrac{ze^{2}}{r}$$

$$\therefore K.E. \dfrac{1}{2}mv^2=\dfrac{Ze^2}{2r}$$
Question 7
1 / -0

A Proton and an $$/alpha $$ particle are accelerated through a potential difference of 100V. The ratio of the wavelength associated with the proton to that associated with an $$/alpha $$ particle is

A
$$1:2$$

B
$$2:1$$

C
$$22:1$$

D
$$2 \sqrt{2} : 1$$

Solution

$$Enrgy=\dfrac{Momentum^2}{2m}=p^2/2m$$
So momentum $$p=\sqrt[2]{2mE}$$, $$m$$ is mass.
Wavelength$$\lambda=\dfrac{h}{p} i.e$$ $$ \lambda \propto \dfrac{1}{p}$$
So required ratio is $$\sqrt[2]{\dfrac{m_aE_a}{m_pE_p}}$$

Mass of alpha particle is $$m_a=4m$$ where $$m=m_p$$ is mass of proton.
Energy of alpha particle $$E_a=q_aV=2e\times100$$
and energy of proton is $$E_p=q_pV=e\times100$$
Putting all the values we get the ratio as $$\sqrt[2]{\dfrac{800me}{100me}}=\dfrac{2\sqrt[2]{2}}{1}$$
Question 8
1 / -0

Electrons accelerated from rest by a potential difference of $$12.75V$$, are bombarded on a mono-atomic hydrogen gas. Possible emission of spectral lines are:-

A
First three Lyman lines, first two Balmer's lines and first Paschen line

B
First three Lyman lines only

C
First two Balmer's lines only

D
None of the above.

Solution

$$ \begin{array}{l} \text { when it get energy of } 12.75 \text { ev it excited to } \\ n=4 \quad E_{4}-E_{1}=12.75 \mathrm{ev} \end{array} $$
$$ \begin{array}{l} \text { then after it comes down, possible emission will } \\ \text { be } \end{array} $$
$$ \begin{array}{l} \text { Line fall on } n=1 \text { ane lyman line }=3 \\ \text { Line fall on } n=2 \text { are balmer line }=2 \\ \text { Line fall on } n=3 \text { are paschen line }=1 \end{array} $$
Question 9
1 / -0

A positroniuim consists of an electron and a positron revolving about their common centre of mass. Calculate the kinetic energy of the electron in ground state:

A
$$1.51\ eV$$

B
$$3.4\ eV$$

C
$$6.8\ eV$$

D
$$13.6\ eV$$

Solution

$$E_k \propto m$$
Where $$m$$ is reduced mass of electron in the orbit.
For hydrogen atom $$m=\dfrac{m_em_p}{m_p +m_e}$$ which is $$m_e$$ .......(1)
because mass of proton $$m_p$$ is much larger
than the mass of electron $$m_e$$.
Mass of positron is equal to mass of electron so reduced mass of electron
for this pair will be $$m=m_e^2/2m_e=m_e/2$$ ..............(2)
As energy for hydrogen atom is $$13.6eV$$
so energy for electron-positron pair will be $$13.6/2=6.8eV$$
Because in this case reduced mass is $$m_e/2$$ i.e. half of that in the case of hydrogen atom.(See equation-2 and equation-1)
Option C is correct.
Question 10
1 / -0

The ground state energy of $$H$$- atom is $$-13.6eV$$. The energy needed to ionise $$H$$- atoms from its second excited state is:

A
$$1.51eV$$

B
$$3.4eV$$

C
$$13.6eV$$

D
$$12.1eV$$

Solution

The energy of $$n^{th}$$ state in the hydrogen state will be $$E_n=-\dfrac{13.6eV}{n^2}$$
so this energy for the second $$excited $$ state i.e. $$n=3$$ will be $$E_3=\dfrac{-13.6eV}{9}=-1.51ev$$
So ionization energy for this state will be $$-(-1.51eV)=1.51eV$$

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Atoms Test - 51

Result

Atoms Test - 51

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Question 1

3

4

5

6

Solution

Question 2

$$r_\alpha = r_p > r_d$$

$$r_\alpha = r_p< r_d$$

$$r_\alpha > r_d > r_p$$

$$r_\alpha = r_d = r_p$$

Solution

Question 3

Lyman series

Balmer's series

Both (1) and (2)

Neither (1) or (2)

Solution

Question 4

$$0.193\ A$$

$$4.24\ A$$

$$2.12\ A$$

$$8.48\ A$$

Solution

Question 5

$${n^2}$$

$${n^4}$$

$${n^6}$$

$${n^8}$$

Solution

Question 6

$$\dfrac{1}{2} \dfrac{Ze^2}{r}$$

$$\dfrac{1}{2}\dfrac{Ze^2}{r^2}$$

$$\dfrac{Ze^2}{r}$$

$$\dfrac{Ze^2}{r^2}$$

Solution

Question 7

$$1:2$$

$$2:1$$

$$22:1$$

$$2 \sqrt{2} : 1$$

Solution

Question 8

First three Lyman lines, first two Balmer's lines and first Paschen line

First three Lyman lines only

First two Balmer's lines only

None of the above.

Solution

Question 9

$$1.51\ eV$$

$$3.4\ eV$$

$$6.8\ eV$$

$$13.6\ eV$$

Solution

Question 10

$$1.51eV$$

$$3.4eV$$

$$13.6eV$$

$$12.1eV$$

Solution

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