Atoms Test

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Atoms Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

The moment of momentum of electrons in the third orbit in a hydrogen atom

A
$$31.67 \times 10^{-34} gm \ m^2/s$$

B
$$3.167 \times 10^{-34} kg \ m^2/s$$

C
$$31.67 \times 10^{-34} kg \ cm^2/s$$

D
$$31.67 \times 10^{-34} gm \ cm^2/s$$

Solution
Question 2
1 / -0

In a sample of hydrogen like atoms all of which are in ground state , a photon beam containing photons of various energies is passed.In absorption spectrum, five dark lies are observed. The number of bright lines in the emission spectrum will be (Assume that all transitions take place)

A
5

B
10

C
15

D
none of these

Solution

Absorption spectrum is observed with a photon in absorbed and a electron jumps to a higher state. Conversely emission spectrum is observed when electron convert to lower state, emitting a photon.
On considering all possible tramition states, thus no. of absorption and emission lines will be same.
Question 3
1 / -0

In a hypothetical atom like that of hydrogen, the mass of the electrons is doubled. Then the energy $$E_0$$ and radius $$r_0$$ of the first Bohr orbit will be ($$a_0$$ = Bohr radius of hydrogen)

A
$$E_0 = -27.2 eV; r_0 = a_0 /2$$

B
$$E_0 = -27.2 eV; r_0 = a_0 $$

C
$$E_0 = -13.6 eV; r_0 = a_0 /2$$

D
$$E_0 = -13.6 eV; r_0 = a_0 $$

Solution
Question 4
1 / -0

The momentum of the Photon of wavelength 6626 NM will be

A
$$10^{-28}kgms^{-1}$$

B
$$10^{-31} kgms^{-1}$$

C
$$10^{-25} kgms^{-1}$$

D
Zero

Solution

We know,

$$p=\dfrac{h}{\lambda}$$

$$p=\dfrac{6.626\times 10^{-34}}{6626\times 10^{-9}}$$

$$p=10^{-28}\ kgm/s$$

Option $$\textbf A$$ is the correct answer
Question 5
1 / -0

In a hydrogen atom, the electron revolves round the proton in a circular orbit of radus $$0.528 \mathring {A}$$ with a speed of $$2.18 \times 10^6$$ m/s. Calculate the centripetal force acting on the electron. (Mass of electron = $$9.1 \times 10^{-31}kg. I \mathring {A}=10^{-10}m)$$

A
$$10.2 \times 10^{-8}$$

B
$$8.2 \times 10^{-8}$$

C
$$9.2 \times 10^{-8}$$

D
$$7.2 \times 10^{-8}$$
Question 6
1 / -0

The wave number of the radiation whose quantum is $$1erg$$ is

A
$$5\times {10}^{15}{Cm}^{-1}$$

B
$$15\times {10}^{5}{cm}^{-1}$$

C
$$1.5\times {10}^{-15}{cm}^{-1}$$

D
$$5\times {10}^{-21}{cm}^{-1}$$

Solution

Energy=1erg

$$\dfrac {hc}{\lambda}=1erg$$

$$\dfrac{1}{\lambda}=wave\ number$$

$$\dfrac{1}{\lambda}=\dfrac{1}{hc}$$

$$u=\dfrac{1}{hc}$$

$$u=\dfrac{1}{6.626×10^{−27}×3×10^10}$$

$$u=\dfrac{100}{19.878}×10^15=5.03×10^{15}cm^{−1}$$
Question 7
1 / -0

The radius of the first orbit of the $$e^-$$ of a hydrogen atom is $$5.3\times 10^{11}m$$, what is the radius of its second orbit.?

A
$$3.12\times 10^{-10}m$$

B
$$2.12\times 10^{-10}m$$

C
$$4.12\times 10^{-10}m$$

D
$$5.12\times 10^{-10}m$$

Solution

Radius $$=\dfrac{x^2}{z}(r_0)$$
$$n=2,z=1$$
$$Radius =\dfrac{4}{1}r_0$$
$$=4\times 5.3\times 10^{-11}m$$
$$21.2\times 10^{-11}$$
$$2.12\times 10^{-10}m$$
Question 8
1 / -0

Using Bohr's formula for energy quantization, the ionisation potential of the ground state of $$Li^{++}$$ atoms is?

A
$$122$$ V

B
$$13.6$$ V

C
$$3.4$$ V

D
$$10.2$$ V

Solution

Ionisation potential of a Hydrogen like atom is given by,
$$IP=\dfrac{-E_{\bot}}{e}=\dfrac{13.6}{1e}\times \dfrac{z^2}{n^2}$$ [$$E_1=$$ Energy of ground state]
For $$Li^{++}, z=3, n=1$$
$$\Rightarrow IP=\bot 3.6\times 3^2=122.4V$$.
Question 9
1 / -0

The lifetime of an electron in the state $$n=2$$ in the hydrogen atom is about

A
$$10\ ns$$

B
$$1000\ ns$$

C
$$1\ ms$$

D
$$10\ ms$$

Solution

The lifetime of an electron in the state $$n=2$$ in the hydrogen atom is about $$10ns$$.
The correct option is A.
Question 10
1 / -0

When $$_{90}Th^{238}$$ changes into $$_{83}Bi^{222}$$, then the number of emitted $$\alpha$$ and $$\beta$$ particles are ?

A
$$8\alpha,7\beta$$

B
$$4\alpha,7\beta$$

C
$$4\alpha,4\beta$$

D
$$4\alpha,1\beta$$

Solution

$$\begin{array}{l}\text { balancing mass no. } \\\qquad 238=222+4 x+0 \\\qquad \begin{array}{l}4 x=16 \\\text { which means } 4 \alpha \text { particle released }\end{array}\end{array}$$
$$\begin{array}{l}\text { balaneing atomie no. } \\\qquad \begin{aligned}90 &=83+2 x+y(-1) \\7 &=8-y \\y &=1 \\\text { So, } & \text { 1 } \beta \text { particle released } \\4 \alpha, 1 \beta & \text { particle released }\end{aligned}\end{array}$$

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Re-Attempt Weekly Quiz Competition

Atoms Test - 52

Result

Atoms Test - 52

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Question 1

$$31.67 \times 10^{-34} gm \ m^2/s$$

$$3.167 \times 10^{-34} kg \ m^2/s$$

$$31.67 \times 10^{-34} kg \ cm^2/s$$

$$31.67 \times 10^{-34} gm \ cm^2/s$$

Solution

Question 2

5

10

15

none of these

Solution

Question 3

$$E_0 = -27.2 eV; r_0 = a_0 /2$$

$$E_0 = -27.2 eV; r_0 = a_0 $$

$$E_0 = -13.6 eV; r_0 = a_0 /2$$

$$E_0 = -13.6 eV; r_0 = a_0 $$

Solution

Question 4

$$10^{-28}kgms^{-1}$$

$$10^{-31} kgms^{-1}$$

$$10^{-25} kgms^{-1}$$

Zero

Solution

Question 5

$$10.2 \times 10^{-8}$$

$$8.2 \times 10^{-8}$$

$$9.2 \times 10^{-8}$$

$$7.2 \times 10^{-8}$$

Question 6

$$5\times {10}^{15}{Cm}^{-1}$$

$$15\times {10}^{5}{cm}^{-1}$$

$$1.5\times {10}^{-15}{cm}^{-1}$$

$$5\times {10}^{-21}{cm}^{-1}$$

Solution

Question 7

$$3.12\times 10^{-10}m$$

$$2.12\times 10^{-10}m$$

$$4.12\times 10^{-10}m$$

$$5.12\times 10^{-10}m$$

Solution

Question 8

$$122$$ V

$$13.6$$ V

$$3.4$$ V

$$10.2$$ V

Solution

Question 9

$$10\ ns$$

$$1000\ ns$$

$$1\ ms$$

$$10\ ms$$

Solution

Question 10

$$8\alpha,7\beta$$

$$4\alpha,7\beta$$

$$4\alpha,4\beta$$

$$4\alpha,1\beta$$

Solution

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