Atoms Test

Result

Atoms Test - 78

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Weekly Quiz Competition

Question 1
1 / -0

The electric potential between a proton and an electron is given by $$V=VIn\dfrac { r }{ { r }_{ 0 } } $$, where $${ r }_{ 0 }$$ is a constant. Assuming Bohr's model to be applicable,write variation of $${ r }_{ n}$$ with $$n,\ n$$ being the principal quantum number.

A
$${ r }_{ n }\propto n$$

B
$${ r }_{ n }\propto \dfrac { 1 }{ n } $$

C
$${ r }_{ n }\propto { n }^{ 2 }$$

D
$${ r }_{ n }\propto \dfrac { 1 }{ { n }^{ 2 } } $$

Solution
Question 2
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Mean free path of gas molecule at constant temperature is inversely proportional to

A
$$P$$

B
$$V$$

C
$$m$$

D
$$n$$ (number density)

Solution
Question 3
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The photon radiated from hydrogen corresponding to $$2nd$$ line of Lyman series is absorbed by a hydrogen like atom $$'X'$$ in $$2nd$$ excited state. As a result the hydrogen like atom $$'X'$$ makes a transition to $$n^{th}$$ orbit. Then,

A
$$X=He^+, n=4$$

B
$$X=Li^{++}, n=6$$

C
$$X=He^+, n=6$$

D
$$X=Li^{++}, n=9$$

Solution
Question 4
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The de Broglie wavelength of an electron in $$n^{th}$$ Bohr orbit of radius $$a_n(a_0$$ being the first orbit radius$$)$$:

A
$$2\pi a_0$$

B
$$2\pi n^2a_0$$

C
$$2\pi na_0$$

D
$$\dfrac{2\pi a_0}{n}$$

Solution
Question 5
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Let $$ F _ { 1 } $$ be the frequency of second line of Lyman series and $$ F _ { 2 } $$ be the frequency of first-line ofBatmer series then frequency of first line of Lyman series is given by

A
$$ F _ { 1 } - F _ { 2 } $$

B
$$ F _ { 1 } + F _ { 2 } $$

C
$$ F _ { 2 } - F _ { 1 } $$

D
$$ \frac { F _ { 1 } F _ { 2 } } { F _ { 1 } + F _ { 2 } } $$

Solution
Question 6
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If $$\lambda_{max.} = 6563 \dot{A}$$ , then wavelength of second line for Balmer series will be :

A
$$\lambda = \dfrac{16}{3 R}$$

B
$$\lambda = \dfrac{36}{5 R}$$

C
$$\lambda = \dfrac{4}{3 R}$$

D
none of these

Solution
Question 7
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The magnetic field at the center of a hydrogen atom due to the motion of electron varies with the principal quantum number as

A
$$n^{5}$$

B
$$n^{3}$$

C
$$1/n^{5}$$

D
$$1/n^{3}$$

Solution
Question 8
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The wavelength of second line of Balmer series is 4815 $$\mathring A $$. The wavelength of first line Balmer series in the same atom is

A

4000 $$\mathring A $$

B
5000 $$\mathring A $$

C
6500$$\mathring A $$

D
7500 $$\mathring A $$

Solution
Question 9
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According to DE Broglie, wavelength of electron in second orbit is 10$$^{ -9 }$$ meter. Then the circumstances of orbit is :-

A
10$$^{ -9 }$$ m

B
2×10$$^{ -9 }$$m

C
3×10$$^{ -9 }$$m

D
4×10$$^{ -9 }$$m

Solution
Question 10
1 / -0

Imagine an atom made of a proton and a hypothetical particle of double the mass as that of an electron but the same charge. Apply Bohr theory to consider transitions of the hypothetical particle to the ground state . Then, the longest wavelength (in terms of Rydberge constant for hydrogen atom ) is

A
$$\dfrac { 1 } { 2 R }$$

B
$$\dfrac { 5} { 3 R }$$

C
$$\dfrac { 1 } { 3 R }$$

D
$$\dfrac { 2 } { 3 R }$$

Solution