Atoms Test

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Atoms Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

If photons of energy $$12.75eV$$ are passing through hydrogen gas in ground state then no of lines in emission spectrum will be

A
$$6$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

$$\begin{array}{l}\text { Given- * Energy of photons }=12.75\mathrm{eV} \\\Rightarrow \text { Let the state of electron initially be } n^{\text {th }} \text { state. } \\\begin{aligned}\Rightarrow \quad 12.75=& 13.6\left[\frac{1}{1}-\frac{1}{n^{2}}\right] \\& \Rightarrow n=4 \\\text { No. of Spectral lines } &=\frac{n(n-1)}{2}=6\end{aligned}\end{array}$$
Question 2
1 / -0

In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy $$E_o$$ and the radius $$r_o$$ of the first orbit will be ($$a_o$$ is the bohr radius)

A
$$E_o=-27.2eV;r_o=a_o/2$$

B
$$E_o=-27.2eV;r_o=a_o$$

C
$$E_o=-13.6eV;r_o=a_o/2$$

D
$$E_o=-13.6eV;r_o=a_o$$

Solution

$$\begin{array}{l}\text { Given-* Mass of the } e^{-}(m) \text { is doubled } \\\text { we know that Energy of an } e^{-} \text {is given by- } \\\qquad E_{0}=-\frac{m e^{4} z^{2}}{8 \varepsilon_{0}^{2} n^{2} h^{2}}=-13.6 \text { ev (at ground state) }\end{array}$$

$$\begin{array}{l}\text { if mass is doubled then - } \\\qquad E_{0}\text { is also double } \left.\Rightarrow [E_{f}=-27.2\mathrm{eV}\right] \\\text { we also know that radius of the } e^{-} \text {is given by (in an orbit) - } \\\text { orbitat radius re }=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m z e^{2}}=a_{0}\text { (bohr radius n=1) } \\\left.\qquad[ r=\frac{a_{0}}{2}\right]\text{If the mass is doubled}\end{array}$$
Question 3
1 / -0

If an electron has, orbital angular momentum quantum number $$l = 7$$, then it will have an orbital angular momentum equal to

A
$$7\left (\dfrac {h}{2\pi}\right )$$

B
$$42\left (\dfrac {h}{2\pi}\right )$$

C
$$2\sqrt {14}\left (\dfrac {h}{2\pi}\right )$$

D
$$\sqrt {56}\left (\dfrac {h}{2\pi}\right )$$

Solution

$$\begin{array}{l}\text { Given- * oribital angular momentum number }\ell=7 \\\text { we know that - } \\\text { orbital angular momentum }=\sqrt{\ell(\ell+1)} \cdot\left(\frac{h}{2 \pi}\right) \\\text { which is equal to }=2 \sqrt{14} \cdot\left(\frac{h}{2 \pi}\right) \text { for } l=7\end{array}$$
Question 4
1 / -0

Consider an atom of $${Li}^{2+}$$ which is a uni-electron system and hence Bohr's model is applicable to it.
If $${r}_{0}=$$ radius of first orbit in hydrogen atom, what is the radius of revolution of electron in second excited state in $${Li}^{2+}$$?

A
$${r}_{0}$$

B
$$2{r}_{0}$$

C
$$3{r}_{0}$$

D
$$9{r}_{0}$$

Solution

$$\text{2nd excited state means 3rd normal state i.e n=3}$$
$$r_n=r_0\dfrac{n^2}{Z}$$ ...........(1)
put in the equation-1 $$Z=3(Lithium), n=3$$ we get $$r_3=3r_0$$
Option C is correct.
Question 5
1 / -0

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

A
$$21.1\times 10^{-11}$$m.

B
$$31.1\times 10^{-11}$$m.

C
$$51.1\times 10^{-11}$$m.

D
$$81.1\times 10^{-11}$$m.

Solution

Formula for the radius of orbit is $$r={ r }_{ 0 }\cdot { n }^{ 2 }$$
when $$n=2$$
$$r=5.3\times { 10 }^{ -11 }\times { 2 }^{ 2 }$$
$$=21.2\times { 10 }^{ -11 }m$$
$$=2.12\times { 10 }^{ -10 }m$$
Question 6
1 / -0

The absorption transition between the first and the fourth energy states of hydrogen atom are $$3$$. The emission transition between these states will be:

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Number of emission states are:

$$ 4\,\,to\,\,3 $$

$$ 4\,\,to\,\,2 $$

$$ 4\,\,to\,\,1 $$

$$ 3\,\,to\,\,2 $$

$$ 3\,\,to\,\,1 $$

$$ 2\,\,to\,\,1 $$

So, number of emission states are $$6$$.
Question 7
1 / -0

The wavelengths involved in the spectrum of deuterium $$\left( ^{2}_1D\right)$$ are slightly different from the hydrogen spectrum, because:

A
Sizes of the two nuclei are different

B
Nuclear forces are different in the two cases

C
Masses of the two nuclei are different

D
Attraction between the electron and the nucleus in different in the two cases

Solution

option C mass of $${ _{ 1 }{ H } }^{ 1 }$$ and $${ _{ 1 }{ D } }^{ 2 }$$ are different hence, the corresponding wavelength are different
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

Bohr’s formula for ionization potential is given by:
$$E=\dfrac{m{{Z}^{2}}{{e}^{4}}}{32{{\pi }^{2}}\varepsilon _{0}^{2}{{h}^{2}}}$$
Where, Z is atomic number = 3
So, $${{E}_{\infty }}-{{E}_{1}}=122\,eV$$
Question 9
1 / -0

If the difference between $$(n + 1)^{th}$$ Bohr radius and $$n^{th}$$ Bohr radius is equal to the $$(n -1)^{th}$$ Bohr radius then find the value of $$n$$

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

Radius of $$n^{th} orbit$$ is $$r_n=n^2r_0$$ where $$r_0$$ is radius of ground state.
Given that $$(n+1)^2r_0 - n^2r_0=(n-1)^2r_0$$
Solving above equation will result in $$n=4$$
Question 10
1 / -0

Energy required for the electron excitation in $$Li^{++}$$ from the first to the third Bohrs orbit:

A
$$7.625\times { 10 }^{ 28 }mev$$

B
$$109 ev$$

C
$$8.8\times { 10 }^{ -28 }mev$$

D
$$56ev$$

Solution

$$\begin{array}{l}\text { we know that energy for exitation of } e^{-}\text {is given as - } \\\qquad E=13.6 \times z^{2}\times\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \\\text { For exitation from first orbit }\left(n_{1}=1\right) \text { to third orbit }\left(n_{2}=3\right) \\\text { we get }-E=13.6\times(3)^{2}\times\left[\frac{1}{1}-\frac{1}{9}\right]=108.8\mathrm{eV} \\E \approx 109 \mathrm{eV} \\\text {Hence, option (B) is correct. }\end{array}$$

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Re-Attempt Weekly Quiz Competition

Atoms Test - 50

Result

Atoms Test - 50

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SHARING IS CARING

Question 1

$$6$$

$$4$$

$$3$$

$$2$$

Solution

Question 2

$$E_o=-27.2eV;r_o=a_o/2$$

$$E_o=-27.2eV;r_o=a_o$$

$$E_o=-13.6eV;r_o=a_o/2$$

$$E_o=-13.6eV;r_o=a_o$$

Solution

Question 3

$$7\left (\dfrac {h}{2\pi}\right )$$

$$42\left (\dfrac {h}{2\pi}\right )$$

$$2\sqrt {14}\left (\dfrac {h}{2\pi}\right )$$

$$\sqrt {56}\left (\dfrac {h}{2\pi}\right )$$

Solution

Question 4

$${r}_{0}$$

$$2{r}_{0}$$

$$3{r}_{0}$$

$$9{r}_{0}$$

Solution

Question 5

$$21.1\times 10^{-11}$$m.

$$31.1\times 10^{-11}$$m.

$$51.1\times 10^{-11}$$m.

$$81.1\times 10^{-11}$$m.

Solution

Question 6

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 7

Sizes of the two nuclei are different

Nuclear forces are different in the two cases

Masses of the two nuclei are different

Attraction between the electron and the nucleus in different in the two cases

Solution

Question 8

$$122 \,V$$

$$13.6 \,V$$

$$3.4 \,V$$

$$10. 2 \,V$$

Solution

Question 9

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 10

$$7.625\times { 10 }^{ 28 }mev$$

$$109 ev$$

$$8.8\times { 10 }^{ -28 }mev$$

$$56ev$$

Solution

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