Atoms Test

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Atoms Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Wavelength of the first line of Balmer series is $$600nm$$. The wavelength of second line of the Balmer series will be:

A
$$444nm$$

B
$$800nm$$

C
$$388nm$$

D
$$632nm$$

Solution

$$\dfrac{1}{\lambda_1} = R \left[\dfrac{1}{2^2} - \dfrac{1}{3^2}\right] = \dfrac{5}{36}R$$
$$\dfrac{1}{\lambda_2} = R\left[\dfrac{1}{2^2} - \dfrac{1}{4^2} \right] = R \left[\dfrac{3}{16}\right]$$
$$\dfrac{\lambda_2}{\lambda_1} = \dfrac{5R/36}{3R/16} = \dfrac{5}{36}\times \dfrac{16}{3}$$
$$\lambda_2 = \dfrac{5}{36}\times \dfrac{16}{3}\times 660 = 444.44nm$$
Question 2
1 / -0

Ionized hydrogen atoms and $$\alpha$$-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths $$r_H : r_{\alpha}$$ will be :

A
$$2 : 1$$

B
$$1 : 2$$

C
$$4 : 1$$

D
$$1 : 4$$

Solution

We know,
$$r_H = \dfrac{P}{e\times B}$$

$$r_{\alpha} = \dfrac{P}{2e\times B}$$(As $$\alpha$$ has 2e of charge on it)

$$\dfrac{r_H}{r_{\alpha}} = \dfrac{\dfrac{P}{eB}}{\dfrac{P}{2eB}}$$

$$\implies \dfrac{r_H}{r_{\alpha}} = \dfrac{2}{1}$$
Hence option A is correct answer
Question 3
1 / -0

The total energy of an electron revolving in the second orbit of hydrogen atom is?

A
$$-13.6$$ eV

B
$$-1.51$$ eV

C
$$-3.4$$ eV

D
Zero

Solution

$$E_n=\dfrac{-13.6z^2}{n^2}$$
$$Z=1; n=2; E_n=-3.4eV$$.
Question 4
1 / -0

The period of revolution of an electron in the ground state of hydrogen atom is T. The period of revolution of the electron in the first excited state is?

A
$$2$$ T

B
$$4$$ T

C
$$6$$ T

D
$$8$$ T

Solution

$$T^2\alpha R^3\alpha n^6\Rightarrow T\alpha n^3$$
Hence $$\dfrac{T^1}{T}=2^3=8$$.
Question 5
1 / -0

In $$H$$-atm spectrum $$V$$ is the wave number
$$V_1 = V_{min} + V_{max}$$ for Lyman series
$$V_1 = V_{min} + V_{max}$$ for Balmer series then $$V_1 : V_2$$

A
$$9:2$$

B
$$3:2$$

C
$$5:2$$

D
$$7:2$$

Solution

$$V_1 = R (1)^2 \left[\dfrac{1}{1^2} - \dfrac{1}{2^2}\right] + R(1)^2 \left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right) = R Z^2\left(\dfrac{7}{4}\right)$$
$$V_2 = R (1)^2 \left[\dfrac{1}{2^2} - \dfrac{1}{3^2}\right] + R(1)^2 \left(\dfrac{1}{2^2} - \dfrac{1}{\infty^2}\right) = R (1)^2\left(\dfrac{7}{18}\right)$$
$$V_1 : V_2 = 9 : 2$$
Question 6
1 / -0

An excited $$He^+$$ ion emits two photons in succession, with wavelengths $$108.5 nm$$ and $$30.4 nm$$, in making a transition to ground state. The quantum number $$n$$, corresponding to its initial excited state is (for photon of wavelength $$\lambda$$, energy $$E = \dfrac{1240 eV}{\lambda (in\, nm)}$$)

A
$$n = 5$$

B
$$n = 4$$

C
$$n = 6$$

D
$$n = 7$$

Solution

$$\dfrac{1}{\lambda} = R \left(\dfrac{1}{m^2} - \dfrac{1}{n^2}\right)z^2$$
$$\dfrac{1}{1085} = R \left(\dfrac{1}{m^2} - \dfrac{1}{n^2}\right)^2$$
$$\dfrac{1}{304} = R \left(\dfrac{1}{1^2} - \dfrac{1}{m^2}\right)2^2$$
$$\therefore m = 2$$
$$\therefore n = 5$$
Question 7
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The number density of molecules of a gas depends on their distance $$r$$ from the origin as, $$n(r) = n_0 e^{-\alpha r^4}$$. Then the total number of molecules is proportional to:

A
$$n_0 \alpha^{1/4}$$

B
$$n_0 \alpha^{-3}$$

C
$$n_0 \alpha^{-3/4}$$

D
$$\sqrt{n_0}\alpha^{1/2}$$

Solution

Given number density of molecules of gas as a function of $$r$$ is

$$n(r) = n_0 e^{-\alpha r^4}$$

$$\therefore$$ Total number of molecule $$= \overset{\infty}{\underset{0}{\int}} n(r) dV$$

$$= \overset{\infty}{\underset{0}{\int}} n_9 e^{-\alpha r^4} 4\pi r^2 dr$$

$$\therefore$$ Number of molecules is proportional to $$n_0 \alpha^{-3/4}$$
Question 8
1 / -0

The ratio of mass densities of nuclei of $$^{40} Ca$$ and $$^{16} O$$ is close to :

A
$$1$$

B
$$2$$

C
$$0.1$$

D
$$5$$

Solution

$$\rho= \dfrac{Mass}{Volume}$$

$$r = r_0 A^{\dfrac {1}{3}}$$
r = radius of nucleus
A = Atomic number
$$r_0 = constant$$

$$\rho = \dfrac{Mass}{\dfrac{4}{3}\pi r^3}$$

$$\rho \propto \dfrac{Mass(Z)}{A}$$
For, Oxygen, Z=16, A=8
Calsium, Z=40, A=20
$$\dfrac{\rho_{o_{2}}}{\rho_{ca}}=\dfrac{Z_{o_{2}}}{Z_ca}=\dfrac{16}{8}\times \dfrac{20}{40} = \dfrac{1}{1}$$
Question 9
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Taking the wavelength of first Balmer line in hydrogen spectrum ($$n=3$$ to $$n=2$$) as $$660nm$$, the wavelength of the 2nd Balmer line ($$n=4$$ to $$n=2$$) will be:

A
$$889.2nm$$

B
$$642.7nm$$

C
$$488.9nm$$

D
$$388.9nm$$

Solution

$$\cfrac { 1 }{ 660 } =R\left( \cfrac { 1 }{ { 2 }^{ 2 } } -\cfrac { 1 }{ { 3 }^{ 2 } } \right) =\cfrac { 5R }{ 36 } ....(1)\quad $$
$$\cfrac { 1 }{ \lambda } =R\left( \cfrac { 1 }{ { 2 }^{ 2 } } -\cfrac { 1 }{ { 4 }^{ 2 } } \right) =\cfrac { 3R }{ 16 } ....(2)\quad \quad $$
dividing (1) with (2)
$$\cfrac { \lambda }{ 660 } =\cfrac { 5\times 16 }{ 36\times 3 } $$
$$\lambda =\cfrac { 4400 }{ 9 } =488.88=488.9nm$$
Question 10
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In $$Li^{++}$$, electron in first Bohr orbit is excited to a level by a radiation of wavelength $$\lambda$$. when the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $$\lambda$$ ?
(Given : $$h = 6.63 \times 10^{34} Js$$; $$ \, c = 3 \times 10^8 ms^{-1}$$)

A
$$9.4 nm$$

B
$$12.3 nm$$

C
$$10.8 nm$$

D
$$11.4 nm$$

Solution

$$\Delta E = \dfrac{hc}{\lambda}$$
$$13.6 \times 9 - 0.85 \times 9 = \dfrac{hc}{\lambda}$$
$$\lambda = \dfrac{hc}{9 \times (13.6 - 0.85) eV}$$
$$= \dfrac{1240 eV . nm}{9 \times 12.75 eV}$$
$$\lambda = 10.8 nm$$

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Re-Attempt Weekly Quiz Competition

Atoms Test - 60

Result

Atoms Test - 60

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SHARING IS CARING

Question 1

$$444nm$$

$$800nm$$

$$388nm$$

$$632nm$$

Solution

Question 2

$$2 : 1$$

$$1 : 2$$

$$4 : 1$$

$$1 : 4$$

Solution

Question 3

$$-13.6$$ eV

$$-1.51$$ eV

$$-3.4$$ eV

Zero

Solution

Question 4

$$2$$ T

$$4$$ T

$$6$$ T

$$8$$ T

Solution

Question 5

$$9:2$$

$$3:2$$

$$5:2$$

$$7:2$$

Solution

Question 6

$$n = 5$$

$$n = 4$$

$$n = 6$$

$$n = 7$$

Solution

Question 7

$$n_0 \alpha^{1/4}$$

$$n_0 \alpha^{-3}$$

$$n_0 \alpha^{-3/4}$$

$$\sqrt{n_0}\alpha^{1/2}$$

Solution

Question 8

$$1$$

$$2$$

$$0.1$$

$$5$$

Solution

Question 9

$$889.2nm$$

$$642.7nm$$

$$488.9nm$$

$$388.9nm$$

Solution

Question 10

$$9.4 nm$$

$$12.3 nm$$

$$10.8 nm$$

$$11.4 nm$$

Solution

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