Atoms Test

Result

Atoms Test - 46

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In Bohr's model of the hydrogen atom, the ratio between the period of revolution of an electron in the orbit of $$n=1$$ to the period of revolution of the electron in the orbit $$n=2$$ is?

A
$$2:1$$

B
$$1:2$$

C
$$1:4$$

D
$$1:8$$

Solution

$$\textbf{Hint:}$$ Use the time period formula and Bohr's velocity and radius formula.

$$\textbf{Step 1: Writing the formulae,}$$
Time period of a revolution is given by,
$$T=\dfrac { 2\pi r }{ V } $$
and we know that radius, r is given by
$$r=\dfrac { { n }^{ 2 }{ h }^{ 2 } }{ 4{ \pi }^{ 2 }mkZ{ e }^{ 2 } } $$
Similarly, velocity v is given by
$$V=\dfrac { 2\pi kZ{ e }^{ 2 } }{ nh } $$

$$\textbf{Step 2:Substituting r and v in formula of time period}$$
$$T=\dfrac{ n^3\,h^3}{4\,\pi^2k^2Z^2e^3m}$$
i.e $$T\propto n^3$$
$$\dfrac{T_1}{T_2}=\left(\dfrac{n_1}{n_2}\right)^3$$
$$\dfrac{T_1}{T_2}=\left(\dfrac{1}{2}\right)^3$$
$$\dfrac{T_1}{T_2}=\left(\dfrac{1}{8}\right)$$
$$\textbf{Option D is correct.}$$
Question 2
1 / -0

A particle of mass m moves around the origin in a potential $$\displaystyle\dfrac{1}{2}mw^2r^2$$, where r is the distance from the origin. Applying the Bohr model in this case, the radius of the particle in its $$n^{th}$$ orbit in terms of $$a=\sqrt{h/(2\pi m\omega)}$$ is?

A
$$a\sqrt{n}$$

B
$$an$$

C
$$an^2$$

D
$$an\sqrt{n}$$

Solution

The force at a distance r is $$F =- \frac{dV}{dr}=me^2r$$
If r is the radius of the n-th orbit, centripetal force is equated to F.
$$\frac{mv^2}{r} = me^2r$$ .........(1)
Using Bohr's quantization, $$mvr = \frac{nh}{2\pi}$$ ....(2)
Solving (1) and (2) by eliminating v, we get $$r = a\sqrt{n}$$
Question 3
1 / -0

The angular momentum of an electron in a hydrogen atom is proportional to(where n is principle quantum number).

A
$$n$$

B
$$n^2$$

C
$$n^3$$

D
$$\sqrt{n}$$

Solution

According to Bohr's model
The angular momentum of orbiting electron =L =mvr = $$\dfrac{nh}{2 \pi}$$
Therefore,
$$L \propto n$$
Question 4
1 / -0

An electron with kinetic energy E collides with a hydrogen atom in the ground state. The collision will be elastic.

A
for all values of E.

B
for $$E < 10.2$$ eV.

C
for $$10.2$$ eV $$<$$ E $$< 13.6$$ eV only.

D
for $$0 <$$ E $$< 3.4$$ eV only.

Solution

$$\begin{array}{l}\text { Energy }=E \\\text { Energy at ground} \text { State for hydrogen like atom}=-13.6\frac{z^{2}}{n^{2}}=V \\\text { }\text {For hydrogen atom, } z=1 \\n=1\end{array}$$

$$E_{1}=-13.6 \times \frac{1^2}{1^{2}}=-13.6 \mathrm{eV}$$

$$\text{for 2nd shell}$$
$$\begin{aligned}E &=-13.6 \times\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right) \\&=-10.2 \mathrm{eV}\end{aligned}$$
$$10.2<E<13.6$$

$$\therefore \text{Opion C is correct}$$
Question 5
1 / -0

Energy levels I, II, III of a certain atom correspond to increasing values of energy i.e., $$E_I < E_{II} < E_{III}$$. If $$\lambda_1$$,$$ \lambda_2$$,$$\lambda_3$$ be the wavelengths of radiations corresponding to the transitions III to II, II to I and III to I respectively, which of the following relations is correct ?

A
$$\lambda_1 = \lambda_2 + \lambda_3$$

B
$$\lambda_3 = \frac {\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$$

C
$$ \lambda_1 + \lambda_2 + \lambda_3 = 0$$

D
$$ \lambda \frac{2}{3} = \lambda \frac{2}{1} + \lambda \frac{2}{2}$$

Solution

The sum of the energies for transitions III to II and II to I is equal to the energy for transition III to I. In other words, the energies and frequencies are additive whereas wavelength is inversely proportional to the frequency. So the reciprocals of the wavelengths are additive.

Thus, $$\cfrac{1}{\lambda_3}=\cfrac{1}{\lambda_2}+\cfrac{1}{\lambda_1}$$
or
$$\lambda_3=\cfrac{\lambda_1\lambda_1}{\lambda_1+\lambda_2}$$
Question 6
1 / -0

In Bohr's model of hydrogen atom, radius of the first orbit of an electron is $$r_0$$. Then, radius of the third orbit is?

A
$$\displaystyle\frac{r_0}{9}$$

B
$$r_0$$

C
$$3r_0$$

D
$$9r_0$$

Solution

Radius of the nth orbit of an electron is given by    $$r_n = r_o\dfrac{n^2}{Z}$$
For hydrogen atom,   $$Z = 1$$
$$\implies \ r_n = r_o n^2$$
So, for third orbit    $$n =3$$
Thus radius of third orbit   $$r_3 = r_o \times 3^2 = 9r_o$$
Question 7
1 / -0

If the first line of Lyman series has a wavelength $$1215.4\mathring { A } $$, the first line of Balmer series is approximately

A
$$4864\mathring { A } $$

B
$$1025.5\mathring { A } $$

C
$$6563\mathring { A } $$

D
$$6400\mathring { A } $$

Solution

Wavelength of first line in Lyman series $$\lambda_L =\dfrac{3}{4R} $$
$$\therefore$$ $$\dfrac{4}{3R} = 1215.4A^o$$
We get $$R = \dfrac{4}{3\times 1215.4} $$
Wavelength of first line in Balmer series $$\lambda_B =\dfrac{36}{5R} $$
$$\implies \ \lambda_B = \dfrac{36\times 3\times 1215.4}{5\times 4} = 6563A^o$$
Question 8
1 / -0

Photons have properties similar to that of.

A
Both particles and waves

B
Particles

C
Waves

D
Neither particles nor waves

Solution

Photons are a type of elementary particle. The quantum of the electromagnetic field including electromagnetic radiation such as light, and the force carrier for the electromagnetic force leven when static via virtual particles.
Question 9
1 / -0

Number of spectral line in hydrogen atom is

A
$$6$$

B
$$8$$

C
$$15$$

D
None of the above

Solution

A
A hydrogen atom has 6 spectral lines. Spectral emissions occurs when an electron transitions jumps from a higher energy state to a lower energy state. The spectral lines are grouped into series according to the lower energy level.
Question 10
1 / -0

If one takes into account finite mass of the proton, the correction to the binding energy of the hydrogen atom is approximately _____.(mass of proton $$1.60\times 10^{-27}\ kg$$, mass of electron $$=9.10\times 10^{-31}\ kg$$).

A
$$0.06\%$$

B
$$0.0006\%$$

C
$$0.02\%$$

D
$$0.00\%$$

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Atoms Test - 46

Result

Atoms Test - 46

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2:1$$

$$1:2$$

$$1:4$$

$$1:8$$

Solution

Question 2

$$a\sqrt{n}$$

$$an$$

$$an^2$$

$$an\sqrt{n}$$

Solution

Question 3

$$n$$

$$n^2$$

$$n^3$$

$$\sqrt{n}$$

Solution

Question 4

for all values of E.

for $$E < 10.2$$ eV.

for $$10.2$$ eV $$<$$ E $$< 13.6$$ eV only.

for $$0 <$$ E $$< 3.4$$ eV only.

Solution

Question 5

$$\lambda_1 = \lambda_2 + \lambda_3$$

$$\lambda_3 = \frac {\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$$

$$ \lambda_1 + \lambda_2 + \lambda_3 = 0$$

$$ \lambda \frac{2}{3} = \lambda \frac{2}{1} + \lambda \frac{2}{2}$$

Solution

Question 6

$$\displaystyle\frac{r_0}{9}$$

$$r_0$$

$$3r_0$$

$$9r_0$$

Solution

Question 7

$$4864\mathring { A } $$

$$1025.5\mathring { A } $$

$$6563\mathring { A } $$

$$6400\mathring { A } $$

Solution

Question 8

Both particles and waves

Particles

Waves

Neither particles nor waves

Solution

Question 9

$$6$$

$$8$$

$$15$$

None of the above

Solution

Question 10

$$0.06\%$$

$$0.0006\%$$

$$0.02\%$$

$$0.00\%$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.