Atoms Test

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Atoms Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

With increasing quantum number, energy difference between adjacent levels in atoms :

A
decreases

B
increases

C
remain constant

D
zero

Solution

$$\begin{array}{l}\text { With increasing quantum}\text { number, energy difference } \\\text { between adjacent levels}\text { in atoms decreases. }\end{array}$$
Question 2
1 / -0

Ratio of minimum to maximum wavelength in Ballmer series is :

A
0.214583333

B
0.233333333

C
1:04

D
3:04

Solution

$$\begin{array}{l}\text { For maximum, the ray should come from the } \\\text { just one level above the Balmer series i.e. } \\\text { Paschenn serios. } \\\text { for minimum the ray should come from } \\\text { infinity to Balmer series. }\end{array}$$
$$\begin{array}{l}\lambda=\text { wavelength } \\R=\text { Rydberg Constant } =1.01\times10^{7}\mathrm{~m}^{-1}\\n=\text { Respective}\text { series }\end{array}$$
$$\begin{array}{l}\text { } \\\text {for Lymen, n }=1\\\text {for Balmer, n }=2 \\\text {for Paschenn, n }=3 \\\text {for Brackett, n }=4 \\\text {for Pfund, n }=5\end{array}$$

$$\begin{array}{l}\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\\text { for maximum } \\\qquad \begin{array}{l}\frac{1}{\lambda_{1}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \\\text { For minimum } \\\frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1_{}^{2}}{\infty^2}\right)=R\frac{1}{2^{2}}\end{array}\end{array}$$
$$\begin{array}{l}\text { Ratio of Minimum to maximum = }\frac{\lambda_{2}}{\lambda_{1}}=0.214583333\end{array}$$
Question 3
1 / -0

Velocity of electron in second Bohr's orbit as compared to velocity in first orbit is :

A
equal

B
one half

C
2 times

D
one fourth

Solution

B
Velocity of electron in nth orbit is given by
$$V_{n} = \frac{Ze^{2}}{2nh}$$
$$Z$$ - atomic number
$$e$$- charge on electron
$$n$$ - no. of orbit
$$h$$ - planks constant.
The velocity of electron in the 2nd Bhors orbit composed to first is one half.
Question 4
1 / -0

A radio transmitter operates at a frequency of 1000 kHz and a power of 66 kW. Then, the number of photons emitted per second is

A
$$1000$$

B
$$10^{20}$$

C
$$10^{12}$$

D
$$10^{29}$$

Solution

$$Number\ of\ photons\ emitted\ per\ second\ = \dfrac{Power}{Energy\ of\ photons}$$
$$ = \dfrac{P}{h\nu} = \dfrac{66\times 1000}{6.6\times10^{-34}\times 10^{6}} = 10^{29}$$
Question 5
1 / -0

Energy of each orbit is

A
changed

B
fixed

C
not same

D
effected

Solution

B
A fixed orbit is the concept in atomic physics, where an electron is considered to remain in a specific orbit, at a fixed distance from an atoms nucleus, for a particular energy level.
Question 6
1 / -0

According to the Bohr's atomic model, the relation between principal quantum number (n) and radius of orbit(r) is?

A
$$r\propto n^2$$

B
$$r\propto \displaystyle\frac{1}{n^2}$$

C
$$r\propto \displaystyle \frac{1}{n}$$

D
$$r\propto n$$

Solution

Electron angular momentum about the nucleus is an integer multiple of $$\displaystyle\frac{h}{2\pi}$$, where h is Planck's constant.
Thus, $$I\omega =mvr=\displaystyle \frac{nh}{2\pi}$$
Hence $$r\propto n$$.
Question 7
1 / -0

Which of the following expressions has the correct units to represent the radius of a hydrogen atom in its ground state?

A
$$\varepsilon_0h^2/\pi m_ee^2$$

B
$$h^2m_ee^2/4\pi \varepsilon_0$$

C
$$m_ec^2/h^2e^4$$

D
$$e^4m_e/8\varepsilon^2_0h^2$$

Solution

Radius of nth orbit of hydrogen atom is given by $$\displaystyle r_n=\frac{n^2}{m_e}\left(\displaystyle\frac{h}{2\pi}\right)^2\left(\displaystyle\frac{4\pi \varepsilon_0}{e^2}\right)$$.
For ground state, $$n=1$$
$$\therefore \displaystyle r_1=\frac{\varepsilon_0h^2}{\pi m_ee^2}$$.
Question 8
1 / -0

Bohr's atomic model is based upon :

A
Einstein's relativistic theory

B
classical theory

C
planks quantum theory

D
both b and c

Solution

D
In 1913, a Danish physicist, Neil Bohr proposed Bohr model of the atom. He suggested that electrons could only have certain classical motions. This model depicts the atom is small, positively charged nucleus surrounded by electrons that travel in circular orbit around the centre. The theory was based on classical theory and quantum theory of radiation.
Question 9
1 / -0

According to Bohr's atomic model, angular momentum of electron in nth orbit is equal to an integral multiple of

A
$$2h/\pi$$

B
$$h/2\pi$$

C
$$h\pi/2$$

D
$$h/\pi$$

Solution

$$\text{ Angular momentum of an electron in the}$$
$$n^{\text {th }}\text{orbit is}$$
$$p=\frac{n h}{2 \pi}$$
$$h \rightarrow \text{planck's constant}$$
$$p=n \times \frac{h}{2 \pi}$$
$$n \rightarrow \text{ineger}$$
$$\therefore P\text{ is integral multiple of $\dfrac{h}{2 \pi}$}$$
Question 10
1 / -0

When a hydrogen atom is raised from the ground state to an excited state.

A
Both K.E. and P.E. increase

B
Both K.E. and P.E. decrease

C
The P.E. decreases and K.E. increases

D
The P.E. increases and K.E. decreases

Solution

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Re-Attempt Weekly Quiz Competition

Atoms Test - 45

Result

Atoms Test - 45

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Rank

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SHARING IS CARING

Question 1

decreases

increases

remain constant

zero

Solution

Question 2

0.214583333

0.233333333

1:04

3:04

Solution

Question 3

equal

one half

2 times

one fourth

Solution

Question 4

$$1000$$

$$10^{20}$$

$$10^{12}$$

$$10^{29}$$

Solution

Question 5

changed

fixed

not same

effected

Solution

Question 6

$$r\propto n^2$$

$$r\propto \displaystyle\frac{1}{n^2}$$

$$r\propto \displaystyle \frac{1}{n}$$

$$r\propto n$$

Solution

Question 7

$$\varepsilon_0h^2/\pi m_ee^2$$

$$h^2m_ee^2/4\pi \varepsilon_0$$

$$m_ec^2/h^2e^4$$

$$e^4m_e/8\varepsilon^2_0h^2$$

Solution

Question 8

Einstein's relativistic theory

classical theory

planks quantum theory

both b and c

Solution

Question 9

$$2h/\pi$$

$$h/2\pi$$

$$h\pi/2$$

$$h/\pi$$

Solution

Question 10

Both K.E. and P.E. increase

Both K.E. and P.E. decrease

The P.E. decreases and K.E. increases

The P.E. increases and K.E. decreases

Solution

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