Atoms Test

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Atoms Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

A diatomic molecule is made of two masses $$m_1$$ and $$m_2$$ which are separated by a distance $$r$$. If we calculate its rotational energy by applying Bohr's rule of angular momentum,its energy will be given by ($$n$$ is an integer)

A
$$\displaystyle \frac{(m_{1}+m_{2})^{2}n^{2}h^{2}}{2m_{1}^{2}m_{2}^{2}r^{2}}$$

B
$$\displaystyle \frac{n^{2}h^{2}}{2(m_{1}+m_{2})r^{2}}$$

C
$$\displaystyle \frac{2n^{2}h^{2}}{(m_{1}+m_{2})r^{2}}$$

D
$$\displaystyle \frac{(m_{1}+m_{2})n^{2}h^{2}}{2m_{1}m_{2}r^{2}}$$

Solution

$$\displaystyle r_1= \frac{m_2}{m_1+m_2}r$$
$$\displaystyle r_2= \frac{m_1}{m_1+m_2}r$$
$$\displaystyle (I_1+I_2)\omega= \frac{nh}{2\pi}$$
$$\displaystyle KE= \frac{1}{2}(I_1+I_2)\omega^2=\frac{(m_{1}+m_{2})n^{2}h^{2}}{2m_{1}m_{2}r^{2}}$$
Question 2
1 / -0

Hydrogen atom is excited from ground state to another state with principal quantum number equal to $$4$$. Then the number of spectral lines in the emission spectra will be.

A
$$2$$

B
$$3$$

C
$$5$$

D
$$6$$

Solution

Total number of spectral lines,
$$N = \dfrac {n(n - 1)}{2}$$
here, $$n$$ is principal quantum number, given $$n = 4$$
$$\Rightarrow N = \dfrac {4(4 - 1)}{2} = 6$$

Hence, the number of spectra lines in the emission spectrum will be $$6$$.
Question 3
1 / -0

The transition from the state $$n = 4$$ to $$n = 3$$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

A
$$2\rightarrow 1$$

B
$$3\rightarrow 2$$

C
$$4\rightarrow 2$$

D
$$5\rightarrow 4$$.

Solution

IR corresponds to least value of $$\displaystyle (\frac{1}{n_1^2}-\frac{1}{n_2^2})$$ i.e. from Paschen, Bracket and Pfund series.
Energy of transition is less only in this case.
Question 4
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The acceleration of an electron in the first orbit of the hydrogen atom $$(n =1)$$ is :

A
$$\dfrac{h^2}{\pi^2m^2r^3}$$

B
$$\dfrac{h^2}{8\pi^2m^2r^3}$$

C
$$\dfrac{h^2}{4\pi m^2r^3}$$

D
$$\dfrac {2\pi^2kme^2Z}{\eta h}$$

Solution

Orbital Angular momentum
$$mvr=\dfrac{nh}{2\pi }$$ n=1

$$mvr=\dfrac{h}{2\pi }$$

$$v=\dfrac{h}{2\pi mr}$$

$$v^{2}=\left ( \dfrac{h}{2\pi mr} \right )^{2}$$

$$a=\dfrac{V^{2}}{r}=\dfrac{h^{2}}{4\pi ^{2}m^{2}r^{3}}$$
Question 5
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An alpha nucleus of energy $$\overline{2}$$ mv bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

A
$$\displaystyle \frac{1}{\mathrm{Z}\mathrm{e}}$$

B
$$\mathrm{v}^{2}$$

C
$$\displaystyle \frac{1}{\mathrm{m}}$$

D
$$\displaystyle \frac{1}{\mathrm{v}^{4}}$$

Solution

Let $$r$$ be the distance of closest approach.
At the closest approach, all the kinetic energy of alpha nucleus has converted into the potential energy i.e. $$K.E = P.E$$
$$\therefore$$ $$\dfrac{1}{2}mv^2 = \dfrac{kZe^2}{r}$$
We get, $$r = \dfrac{2kZe^2}{mv^2}$$
$$\implies r\propto \dfrac{1}{m}$$
Question 6
1 / -0

The time by a photo-electron to come out after the photon strikes is approximately

A
$$10^{-1}s$$

B
$$10^{-4}s$$

C
$$10^{-10}s$$

D
$$10^{-16}s$$

Solution

From experimental aspect, the photoelectric effect is an instantaneous phenomenon. The photon strikes the metallic surface is approximately $$10^{-10}$$ s.
Question 7
1 / -0

According to Bohr's theory, the time averaged magnetic field at the centre (i.e nucleus) of a bydrogen atom due to the motion of electrons in the $$n^{th}$$ orbit is propotional to : $$(n =$$ principal quantum number$$)$$

A
$$n^{-3}$$

B
$$n^{-4}$$

C
$$n^{-5}$$

D
$$n^{-2}$$

Solution

The magnetic field induced at the centre of an orbit due to an electron revolving in the nth orbit of hydrogen atom is given by
$$B=\mu_0i_{n/(2r_n)}$$,
Where $$i_n$$ is the current in the nth orbit $$r_n$$ the nth orbit and $$\mu_0$$ the magnetic permeablity of free space.
$$r_n\ \alpha\ n^2$$
$$i_n\ \alpha\ n^{-3}$$
Therefore
$$\beta\ \alpha\ n^{-5}$$
So the answer is (c) $$n^{-5}$$.
Question 8
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If one were to apply Bohr model to a particle '$$m$$' and charge '$$q$$' moving in a plane under the influence of a magnetic field 'B', the energy of the charged particle in the $$ n^{th}$$ level will be.

A
$$n\left ( \dfrac {hqB}{4 \pi m} \right )$$

B
$$n\left ( \dfrac {hqB}{8 \pi m} \right )$$

C
$$n\left ( \dfrac {hqB}{\pi m} \right )$$

D
$$n\left ( \dfrac {hqB}{2 \pi m} \right )$$

Solution

According to Bohr model
$$\vec{L}=\dfrac{nh}{2\pi}$$
$$mvr=\dfrac{nh}{2\pi}$$
$$vr=\dfrac{nh}{2m \pi}$$...(i)
$$\dfrac{mv^2}{r}=qvB$$
$$\dfrac{v}{r}=\dfrac{qB}{m}$$...(ii)
multiplying (i) and (ii)
$$\dfrac{v}{r}=\dfrac{qB}{m}$$
$$v^2=\dfrac{nhqB}{2 \pi m^2}$$
K.E.=$$\dfrac{1}{2}mv^2=\dfrac{nhqB}{4 \pi m}$$
The potential energy due to magnetic field will be zero as magnetic field is always perpendicular to velocity, hence work done by magnetic force will be zero.
Question 9
1 / -0

The time period of revolution of electron in its ground state orbit in a hydrogen atom is $$1.6\times 10^{-16}$$ s. The frequency of revolution of the electron in its first excited state (in $$s^{-1}$$) is:

A
$$5.6\times 10^{12}$$

B
$$1.6\times 10^{14}$$

C
$$6.2\times 10^{15}$$

D
$$7.8\times 10^{14}$$

Solution

Angular momentum of the electron is given by
$$L_n=\dfrac{nh}{2\pi}$$, where $$n$$ is the principle quantum

For the ground state, $$n=1$$
$$L_1=\dfrac{h}{2\pi}=mv_1r_1$$

For the first excited state, $$n=2$$
$$L_2=\dfrac{2h}{2\pi}=\dfrac{h}{\pi}mv_2r_2=2\left(\dfrac{h}{2\pi}\right)$$
$$=2L_1$$
$$\therefore L_2=2L_1$$
$$mv_2r_2=2mv_1r_1$$
$$v_2r_2=2v_1r_1$$
$$w_2r^2_2=2w_1r^2_1$$

$$w_2=\dfrac{2w_1r^2_1}{r^2_2},$$ or $$f_2=\dfrac{2f_1r^2_1}{r^2_2}=\dfrac{2r^2_1}{r^2_2T_1}$$

Now, the radius of orbit in a hydrogen is given by
$$r=\dfrac{r_1n^2}{z}=r_1n^2$$
For the first excited state, $$r_2=r_1(2)^2=4r_1$$
$$\therefore \dfrac{r_2}{r_1}=4$$

time period of revolution of electron in its ground state orbit in a hydrogen atom is $$1.6\times 10^{-16}$$ sec

$$\therefore f_2=\dfrac{2r^2_1}{r^2_2 T_1}=\dfrac{2}{16}\times \dfrac{1}{1.6\times 10^{-16}}$$

$$=\dfrac{2\times 10^{16}}{1.6\times 16}=7.812\times 10^{14}$$

option (D) is the answer

Question 10

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Match List - I (Experiment performed) with List - II (Phenomena discovered/ associated) and select the correct option from the options given below the lists :

	List - I		List - II
(a)	Davisson and Germer Experiment	(i)	Wave nature of electrons
(b)	Millikan's oil drop experiment	(ii)	Charge of an electron
(c)	Rutherford experiment	(iii)	Quantisation of energy levels
(d)	Franck - Hertz experiment	(iv)	Existence of nucleus

(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Solution

Davisson and Germer Experiment$$\Rightarrow$$Wave nature of electrons
Millikan's oil drop experiment$$\Rightarrow$$Charge of an electron
Rutherford experiment$$\Rightarrow$$Existence of nucleus
Franck - Hertz experiment$$\Rightarrow$$Quantisation of energy levels .

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Atoms Test - 10

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Atoms Test - 10

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Question 1

$$\displaystyle \frac{(m_{1}+m_{2})^{2}n^{2}h^{2}}{2m_{1}^{2}m_{2}^{2}r^{2}}$$

$$\displaystyle \frac{n^{2}h^{2}}{2(m_{1}+m_{2})r^{2}}$$

$$\displaystyle \frac{2n^{2}h^{2}}{(m_{1}+m_{2})r^{2}}$$

$$\displaystyle \frac{(m_{1}+m_{2})n^{2}h^{2}}{2m_{1}m_{2}r^{2}}$$

Solution

Question 2

$$2$$

$$3$$

$$5$$

$$6$$

Solution

Question 3

$$2\rightarrow 1$$

$$3\rightarrow 2$$

$$4\rightarrow 2$$

$$5\rightarrow 4$$.

Solution

Question 4

$$\dfrac{h^2}{\pi^2m^2r^3}$$

$$\dfrac{h^2}{8\pi^2m^2r^3}$$

$$\dfrac{h^2}{4\pi m^2r^3}$$

$$\dfrac {2\pi^2kme^2Z}{\eta h}$$

Solution

Question 5

$$\displaystyle \frac{1}{\mathrm{Z}\mathrm{e}}$$

$$\mathrm{v}^{2}$$

$$\displaystyle \frac{1}{\mathrm{m}}$$

$$\displaystyle \frac{1}{\mathrm{v}^{4}}$$

Solution

Question 6

$$10^{-1}s$$

$$10^{-4}s$$

$$10^{-10}s$$

$$10^{-16}s$$

Solution

Question 7

$$n^{-3}$$

$$n^{-4}$$

$$n^{-5}$$

$$n^{-2}$$

Solution

Question 8

$$n\left ( \dfrac {hqB}{4 \pi m} \right )$$

$$n\left ( \dfrac {hqB}{8 \pi m} \right )$$

$$n\left ( \dfrac {hqB}{\pi m} \right )$$

$$n\left ( \dfrac {hqB}{2 \pi m} \right )$$

Solution

Question 9

$$5.6\times 10^{12}$$

$$1.6\times 10^{14}$$

$$6.2\times 10^{15}$$

$$7.8\times 10^{14}$$

Solution

Question 10

(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Solution

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