Structure of Atom Test

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Structure of Atom Test - 60

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Question 1
1 / -0

A stationary $$He^+$$ ion emitted a photon corresponding to the first line of the Lyman series. The photon liberates a photo electron from a stationary hydrogen atom in ground state. The kinetic energy?

A
$$10.2 eV$$

B
$$13.6 eV$$

C
$$20.4 eV$$

D
$$27.2 eV$$

Solution
Question 2
1 / -0

If the velocity of an electron moving in first Bohr's orbit of hydrogen has to be reduced to$$7.27\times 10^{5}ms^{-1}$$ it should be shifted from its existing position by

A
$$4.23\mathring{A}$$ away from nucleus

B
$$4.23\mathring{A}$$ towards nucleus

C
$$4.76\mathring{A}$$ away from nucleus

D
$$4.76\mathring{A}$$ towards nucleus

Solution
Question 3
1 / -0

In an electronic transition atom cannot emit -

A
visible light

B
$${ \gamma }$$- rays

C
Infra red light

D
Ultra violet light

Solution
Question 4
1 / -0

Imagine an atom made up of proton and a a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength $$\lambda $$ (give in term of the Rydberg constant R for the hydrogen atom) equal to

A
9/5 R

B
36/5 R

C
18/5 R

D
4/ R

Solution

Energy is related to mass
$${R_n} \propto m$$
The longest wavelength λmax photon will correspond to the transition of particle from$$n=3$$ to$$ n=2$$
$$\begin{array}{l} \dfrac { 1 }{ { { \lambda _{ \max } } } } =2R\left( { \dfrac { 1 }{ { { 2^{ 2 } } } } -\dfrac { 1 }{ { { 3^{ 2 } } } } } \right) \\ { \lambda _{ \max } }=\dfrac { { 18 } }{ 5 } R \end{array}$$
Hence, Option $$C$$ is correct.
Question 5
1 / -0

In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr's orbit in atom of hydrogen

A
$$4\rightarrow 1$$

B
$$2\rightarrow 5$$

C
$$3\rightarrow 2$$

D
$$5\rightarrow 2$$

Solution

We see red end means it is a part of the visible region and obviously only Balmer series corresponds to the visible region for the Balmer series $$n_1=2$$ and red end means low energy or third end from this means $$n_2=-5 $$
$$2(r)5. 3$$
$$2(r)4. 2 $$
$$2(r)3. 1$$
The correct answer is (D) $$5 \to 2$$
Question 6
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In a Bohr's model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be emitted?

A
$$2.15\times10^{-11}$$ ergs

B
$$2.389\times10^{-12}$$ ergs

C
$$0.239\times10^{-10}$$ ergs

D
$$0.1936\times10^{-10}$$ ergs

Solution
Question 7
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The distance between $$3^{rd}$$ & $$2^{nd}$$ Bohr orbit $$He^{+}$$ is

A
$$2.645\ A^o$$

B
$$1.322\ A^o$$

C
$$0.2645\ A^{o}$$

D
$$None\ of\ these$$

Solution

Dis\tance between the $${ 2^{ nd } }$$ and the $${ 3^{ rd } }$$ orbits=distance of $${ 3^{ rd } }$$ orbit-distance of second orbit from the nucleus
$${ d_{ 3 } } = 0.529 \times \dfrac { n^{ 2 }}{ z } \times { 10^{ -10 } } m=0.529 \times 9 \times { 10^{ -10 } } m $$
$${ d_{ 2 } }=0.529\times 4\times { 10^{ -10 } }\, m \\ { d_{ 3 } }-{ d_{ 2 } }=0.529\times 5\times { 10^{ -10 } }\, m =2.645\times { 10^{ -10 } }\, m=2.645A^{ 0 } $$

Hence,
option $$(A)$$ is correct answer.
Question 8
1 / -0

The ratio of the radii of the first three Bohr orbit in H atom is

A
$$1:\dfrac { 1 }{ 2 } :\dfrac { 1 }{ 3 } $$

B
$$1:2:3$$

C
$$1:4:9$$

D
$$1:8:27$$

Solution

Atomic number $$Z$$ is equal to $$1$$.
Hence the radius of $$n th$$ orbit, $$r_n=0.529 n^2 A$$
For first three orbits, $$n$$ values are $$1,2$$ and $$3$$
ratio of radii of first three orbit $${r_1}:{r_2}:{r_3} = n_1^2:n_2^2:n_3^2$$
$$\begin{array}{l} \Rightarrow { 1^{ 2 } }:{ 2^{ 2 } }:{ 3^{ 2 } } \\ \Rightarrow 1:4:9 \end{array}$$
Question 9
1 / -0

The first three radius ratio of Bohr orbits

A
$$1:0.5:0.5$$

B
$$1:2:3$$

C
$$1:4:9$$

D
$$1:8:27$$

Solution
Question 10
1 / -0

According to Bohr's model, if the kinetic energy of an electron in $$2^{nd}$$ orbit of $$He^{+}$$ is $$x$$, then what should be the ionisaion energy of the electron revolving in $$3^{rd}$$ orbit of $$m^{5+} ions>$$

A
$$X$$

B
$$4X$$

C
$$x/4$$

D
$$2X$$

Solution

$$\begin{array}{l} X=13\cdot 6\frac { { { { \left( 2 \right) }^{ 2 } } } }{ { { { \left( 2 \right) }^{ 2 } } } } \\ =13\cdot 6 \\ X'=13\cdot \frac { { { { \left( 6 \right) }^{ 2 } } } }{ { { { \left( 3 \right) }^{ 2 } } } } \\ =13\cdot 6\times \frac { { 36 } }{ 9 } \\ =4X \\ Hence,\, option\, B\, is\, correct\, answer. \end{array}$$