Atoms Test

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Atoms Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

The main defect of Bohr's atom model is :

A
mixing of classical and quantum theories

B
exclusion of nuclear motion

C
failed to explain the fine structure of spectral lines

D
failed to explain larger atoms

Solution

Bohr's model can only clearly explain hydrogen or hydrogen-like atoms, it fails when applied to larger and heavier atoms like iron, gold, mercury, etc.
Question 2
1 / -0

The electron is present in an orbit of energy state $$-1.51$$ eV, then angular momentum of the electron is

A
$$2h/\pi $$

B
$$h/\pi $$

C
$$3h/2\pi $$

D
$$7h/\pi $$

Solution

Given : $$E_n =-1.51$$ eV
Using $$E_n = \dfrac{-13.6}{n^2}$$ eV (for H-atom)

$$\therefore$$ $$-1.51 = \dfrac{-13.6}{n^2}$$ $$\implies n =3$$
Thus angular momentum $$L = \dfrac{nh}{2\pi} = \dfrac{3h}{2\pi}$$
Question 3
1 / -0

The possible values of principal quantum number can be:

A
$$1, 2, 3...8$$

B
$$0, 1, 2...8$$

C
only zero

D
only odd numbers

Solution

Values of Principle quantum number are $$1,2,3,4.....8$$. $$0$$ is not a Principle quantum number.
Only odd numbers are not Principle quantum numbers either.
Odd numbers, as well as even numbers, are Principle quantum number except $$0$$ and negative integers.
Question 4
1 / -0

The radius of shortest orbit in one electron system is $$18$$ pm.It may be.

A
$$_{1}^{1}H$$

B
$$_{2}^{1}H$$

C
$$He^{+}$$

D
$$Li^{+}$$

Solution

Hint:

For an atom of atomic number Z, The radius of $${n^{th}}$$ orbit is given as,

$${r_n} = \dfrac{{{n^2}{a_o}}}{Z}$$

Where, $${a_o}$$ is Bohr’s radius, $${a_o} = 53pm$$

Correct Option : Option D.
Explanation for correct answer:
a)      $${_1^1}H$$

$$Z = 1$$

For shortest orbit, $$n = 1$$

Therefore, $${r_1} = \dfrac{{{1^2} \times 53}}{1} = 53pm$$

b)      $${_2^1}H$$

$$Z = 1$$

For shortest orbit, $$n = 1$$

Therefore, $${r_1} = \dfrac{{{1^2} \times 53}}{1} = 53pm$$

c)      $$H{e^ + }$$

$$Z = 2$$

For shortest orbit, $$n = 1$$

Therefore, $${r_1} = \dfrac{{{1^2} \times 53}}{2} = 26.5pm$$

d)      $$L{i^ + }$$

$$Z = 3$$

For shortest orbit, $$n = 1$$

Therefore, $${r_1} = \dfrac{{{1^2} \times 53}}{3} = 17.667pm$$

The radius of shortest orbit of $$L{i^ + }$$ is closest to $$18pm$$. Thus Option D is the correct answer.
Question 5
1 / -0

Atomic hydrogen is excited to the nth energy level.The maximum number of spectral lines which it can emit while returing to the ground state, is

A
$$\frac{1}{2}n(n-1)$$

B
$$\frac{1}{2}n(n+1)$$

C
$$n(n+1)$$

D
$$n(n-1)$$

Solution

Maximum no of spectral lines $$=\dfrac{n(n-1)}{2}$$
Question 6
1 / -0

If $$A_{n}$$ is the area enclosed in the nth orbit in a hydrogen atom then the graph log $$\left ( \dfrac{A_{n}}{A_{1}} \right )$$ against log $$n$$

A
will have slope $$2$$ (straightline)

B
will have slope $$4$$ (straightline)

C
will be a monotonically increasing non linear curve

D
will be a circle

Solution

Radius of $$n^{th}$$ orbit $$=a_{0}n^{2}$$
Radius of 1st orbit $$=a_{0}$$
Area of $$n^{th}$$ orbit $$=\pi a{_{0}}^{2}n^{4}=A_{n}$$
Area of 1st orbit $$=\pi a{_{0}}^{2}A_{1}$$
$$\dfrac{A_{n}}{A_{1}}=n^{4}\left [ Taking\ log\ both\ sides \right ]$$
$$\Rightarrow log\left ( \dfrac{A_{n}}{A_{1}} \right )=4logn$$

Slope of the above line $$=4$$
Question 7
1 / -0

If an electron is revolving round the hydrogen nucleus at a distance of $$0.1\ nm$$, the speed should be :

A
$$2.188\times 10^{6}m/s$$

B
$$1.094\times 10^{6}m/s$$

C
$$4.376\times 10^{6}m/s$$

D
$$1.59\times 10^{6}m/s$$

Solution

Radius of orbit $$=0.1nm$$
According to formula:
$$0.1\times 10^{-9}=0.529\times 10^{-10}\times n^{2}$$

$$\Rightarrow n^{2}=1.8\Rightarrow n=1.37$$

So, in 2nd case:
Velocity of electron $$=\dfrac{2.18\times 10^{6}}{n}$$

$$\Rightarrow V=\dfrac{2.18}{1.37}\times 10^{6}=1.59\times 10^{6}m/s$$
Question 8
1 / -0

The $$\alpha$$-particles with initial kinetic energy $$4.8$$ mev are shot at gold atoms $$(z=79)$$. The potential of $$\alpha$$-particle at the closest distance to the gold nucleus is

A
$$4.8 MV$$

B
$$zero$$

C
$$9.6 MV$$

D
$$2.4 MV$$

Solution

The energy of alpha particle is $$E = 4.8 MeV$$
Also, for alpha particles ($$e = 2e$$)
$$E = 2eV$$

$$V = \dfrac{E}{2e}$$

$$V = \dfrac{4.8 \times 10^6 \times 1.6 \times 10^{-19}}{2(1.6 \times 10^{-19})}$$

$$V = \dfrac{4.8 \times 10^6}{2} = 2.4 \times 10^6 = 2.4 MV$$
Question 9
1 / -0

When an electron jumps from higher orbit to the second orbit in hydrogen, the radiation emitted out will be in $$(R=1.09\times 10^{7}m^{-1})$$

A
ultraviolet

B
visible region

C
infrared region

D
X-ray region

Solution
Question 10
1 / -0

The maximum number of photons emitted by an H-atom, if atom is excited to states with principal quantum number four is

A
$$4$$

B
$$6$$

C
$$2$$

D
$$1$$

Solution

Hint:

Number of Photons emitted by H-atom if it is excited to $${n^{th}}$$ excited state is,

$$\bf{photons = \dfrac{{n(n - 1)}}{2}}$$

Step 1: Calculate the number of photons emitted by H-atom
Hydrogen atom is excited to $${4^{th}}$$ excited state, so number of photons emitted by using formula is,

$$photons = \dfrac{{n(n - 1)}}{2}$$

$$ \Rightarrow photons = \dfrac{{4(4 - 1)}}{2}$$

$$ \Rightarrow photons = 6$$

Thus, 6 photons are emitted if the Hydrogen atom is excited to $${4^{th}}$$ excited state.

Option B is correct.

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Atoms Test - 26

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Atoms Test - 26

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SHARING IS CARING

Question 1

mixing of classical and quantum theories

exclusion of nuclear motion

failed to explain the fine structure of spectral lines

failed to explain larger atoms

Solution

Question 2

$$2h/\pi $$

$$h/\pi $$

$$3h/2\pi $$

$$7h/\pi $$

Solution

Question 3

$$1, 2, 3...8$$

$$0, 1, 2...8$$

only zero

only odd numbers

Solution

Question 4

$$_{1}^{1}H$$

$$_{2}^{1}H$$

$$He^{+}$$

$$Li^{+}$$

Solution

Question 5

$$\frac{1}{2}n(n-1)$$

$$\frac{1}{2}n(n+1)$$

$$n(n+1)$$

$$n(n-1)$$

Solution

Question 6

will have slope $$2$$ (straightline)

will have slope $$4$$ (straightline)

will be a monotonically increasing non linear curve

will be a circle

Solution

Question 7

$$2.188\times 10^{6}m/s$$

$$1.094\times 10^{6}m/s$$

$$4.376\times 10^{6}m/s$$

$$1.59\times 10^{6}m/s$$

Solution

Question 8

$$4.8 MV$$

$$zero$$

$$9.6 MV$$

$$2.4 MV$$

Solution

Question 9

ultraviolet

visible region

infrared region

X-ray region

Solution

Question 10

$$4$$

$$6$$

$$2$$

$$1$$

Solution

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