Atoms Test

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Atoms Test - 73

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Weekly Quiz Competition

Question 1
1 / -0

In a hypothetical system a particle of mass $$m$$ and charge $$-3q$$ is moving around a very heavy particle having charge $$q$$. Assuming Bohr's model to be true to this system, the orbital velocity of mass $$m$$ when it is nearest to the heavy particle is

A
$$\cfrac{3{q}^{2}}{2{\epsilon}_{0}h}$$

B
$$\cfrac{3{q}^{2}}{4{\epsilon}_{0}h}$$

C
$$\cfrac{3{q}^{}}{2{\epsilon}_{0}h}$$

D
$$\cfrac{3{q}^{}}{4{\epsilon}_{0}h}$$
Question 2
1 / -0

Angular momentum in second Bohr orbit of H-atom is $$x$$. Then find out angular momentum in 1st excited state of $${Li}^{+2}$$ ion:

A
$$3x$$

B
$$9x$$

C
$$\cfrac{x}{2}$$

D
$$x$$

Solution

$$ \begin{array}{l} \text { we know, } \\ \text { Angular momentum }=\frac{n h}{2 \pi} \end{array} $$
$$ \begin{array}{l} \text { Given, } \\ \text { For } n=2 \\ \text { Angular momentum = } x \\ \qquad x=\frac{2 h}{2 \pi} --(i) \end{array} $$
$$ \text { for } 1^{\text {st }} \text { excited state means } n=2 $$
$$ \begin{aligned} & A \cdot M=\frac{2 h}{2 \pi} \quad-(11) \\ \therefore \quad &(i)=(i i) \end{aligned} $$
$$ \text { Angular mementum }= x $$
Question 3
1 / -0

In a H-atom, the transition takes place from L to K shell. If $$R=1.08\times {10}^{7}{m}^{-1}$$, the wave length of the light emitted is nearly

A
$$4400\mathring{A}$$

B
$$1250\mathring{A}$$

C
$$1650\mathring{A}$$

D
$$1850\mathring{A}$$

Solution
Question 4
1 / -0

A positronium consists of an electron and a positron revolving about their common centre of mass. Calculate the separation between the electron and positron in their first excited state:

A
$$0.529\ A$$

B
$$1.058\ A$$

C
$$2.116\ A$$

D
$$4.232\ A$$

Solution

$$\begin{array}{l}\text { Given, positronium consist of electron and positron }\end{array}$$
$$\begin{array}{l}\text { Let w be angular velocity about } 0 . \\\text { balancing force } \\\qquad \frac{k e^{2}}{4 R^{2}}=m \omega^{2} R\text { --- (1) }\end{array}$$
$$\begin{array}{l}\text { Now, Angular momentum } \\\qquad \begin{aligned}m(\omega R) R&=\frac{n h}{2 \pi} \\m \omega R^{2} &=\frac{n h}{2 \pi } ---(2)\end{aligned}\end{array}$$
$$\begin{array}{l}\text { from }(1) \text { and }(2) \\\quad 2 R=\frac{n^{2} h^{2}}{\pi^{2} m K e^{2}}\end{array}$$
$$\begin{array}{l}2 R=4 \times\left(\frac{n^{2} h^{2}}{4 \pi^{2} m K e^{2}}\right) \\2 R=4\times\frac{0.529 n^{2}}{z}\end{array}$$
$$\begin{aligned}&=4 \times 0.529 \dot {A} \quad, \quad n=1,z=1 \\2 R &=4.232 \dot{A}\end{aligned}$$
Question 5
1 / -0

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is :
(in $$A^o$$)

A
1215

B
1640

C
2430

D
4687

Solution
Question 6
1 / -0

Assuming Bohr's model for $$ Li^{++} $$ atom, the first excitatio energy of ground state of $$ Li^{++} $$ atom is

A
10.2 eV

B
91.8 eV

C
13.6 eV

D
3.4 eV

Solution

For $$Li^{++}$$ model,
$$2=3$$
In first excitation of ground state,
$$n_1=1$$ and $$n_2=2$$
In Bohr's model,
$$\Delta E=(13.6ev)2^2\left[\dfrac{1}{n_1^2}-\dfrac{1}{n^2_2}\right]$$
Here, $$\Delta E=(13.6)(3)^2\left[\dfrac{1}{1}-\dfrac{1}{4}\right]$$
$$\Delta E=(13.6)9\times \dfrac{3}{4}=91.8$$eV
So, first excitation energy for ground state in $$Li^{++}$$ is $$91.8$$eV
Option B is correct.
Question 7
1 / -0

Calculate number of electrons present in $$500\ cm^{3}$$ volume of water.

A
$$1.67\times 10^{20}$$

B
$$1.67\times 10^{26}$$

C
$$1.67\times 10^{23}$$

D
$$3.2\times 10^{26}$$

Solution

$$\begin{array}{l}\text { Given } \\\text { of electron in one molecule of }\mathrm{H}_{2}O=2+8 \\=10\end{array}$$
$$\begin{array}{l}\text { we know } \\\text { density of watere } 1 \mathrm{~g} /\mathrm{cm}^{3} \\\text { given volume }=500\mathrm{~cm}^{3} \\\text { No. of mole of water }=\left(\frac{500}{18}\right)\end{array}$$
$$\begin{array}{l}\text { Since } 1 \text { mole contain } 6.02 \times 10^{23} \text { molecule of } H_{2}O \\\text { and } 1 \mathrm { H}_{2}\mathrm{O} \text { contain } 10\mathrm{e}^{-}\end{array}$$
$$\begin{aligned}\qquad \begin{aligned}\text { total } e^{-} \text { contain } &=\frac{500}{18}\times 6.022 \times 10^{23}\times 10 \\&=1.67 \times 10^{26}\end{aligned}\end{aligned}$$
Question 8
1 / -0

The magnitude of angular momentum, orbit radius and frequency of revolution of electron in hydrogen atom corresponding to quantum number $$n$$ are $$L,r$$ and $$f$$ respectivey. Then according to Bohr's theory of hydrogen atom

A
$$f{r}^{2}L$$ is constant for all orbits

B
$$f{r}^{}L$$ is constant for all orbits

C
$${f}^{2}rL$$ is constant for all orbits

D
$$fr{L}^{2}$$ is constant for all orbits

Solution
Question 9
1 / -0

The wavelength of the first member of the Balmer series in hydrogen spectrum is $$x$$ $$\mathring{A}$$. Then the wavelength (in $$\mathring{A}$$) of the first member of Lyman series in the same spectrum is

A
$$\cfrac{5}{27}x$$

B
$$\cfrac{4}{3}x$$

C
$$\cfrac{27}{5}x$$

D
$$\cfrac{5}{36}x$$

Solution

$$ \begin{array}{l} \text { Balmer series : transition take place from } \\ n=2 \text { to } n=3,4,5, \text { so on } \\ \text { for first member of balmer } \\ n_{1}=2 \text { to } n_{2}=3 \end{array} $$
$$ \lambda=x \dot {A} $$
$$ \begin{array}{l} \frac{1}{\lambda}=R z^{2}\left(\frac{1}{n_{1} 2}-\frac{1}{n_{2}^{2}}\right) \\ \frac{1}{x}=R z^{2}\left(\frac{1}{4}-\frac{1}{9}\right) \\ \frac{1}{x}=R z^{2}\left(\frac{5}{36}\right)-(1) \end{array} $$
$$ \begin{array}{l} \text { For first member of Lyman } \\ \qquad \begin{aligned} n_{1} &=3 \quad n_{2}=2 \\ \frac{1}{\lambda} &=R_{2}^{2}\left(\frac{1}{1}-\frac{1}{4}\right) \\ \frac{1}{\lambda} &=R z^{2}\left(\frac{3}{4}\right) \end{aligned} \end{array} $$
$$ \begin{array}{l} \text { divide (i) } \div \text { (ii) } \\ \qquad \begin{array}{l} \frac{ \lambda }{x}=\left(\frac{5}{36}\right)\left(\frac{4}{3}\right) \\ \lambda=\left(\frac{5}{27}\right) x \end{array} \end{array} $$
Question 10
1 / -0

In a hypothetical system a particle of mass m and charge -3q is moving around a very heavy particle having charge q. Assuming Bohr's model to be true to this system, The orbital velocity of mass m when it is nearest to heavy particle is

A
$$\dfrac{3q^2}{2 \varepsilon_0h}$$

B
$$\dfrac{4q^2}{2 \varepsilon_0h}$$

C
$$\dfrac{3q}{2 \varepsilon_0 h}$$

D
$$\dfrac{3q}{4 \varepsilon_0 h}$$

Solution

$$\begin{array}{l}\text { Given, } \\\text { a hypothetical system } \\\text { Bohr's model is true by } \\\text { assuming. }\end{array}$$
$$\text { } \begin{aligned}I) F_{\text {centripetal }} &=\text { culoumb force } \\&\frac{m v^{2}}{r}=\frac{1 \times q \times+3 q}{4 \pi \varepsilon_{0} \times r^{2}} \\& m v r=\frac{3 q^{2}}{4 \pi \varepsilon_{0} v}\cdots(1)\end{aligned}$$
$$\begin{array}{l}\text { II) Angular momentum }=\frac{n h}{2 \pi} \\m v r=\frac{n h}{2\pi}\text { (ii) } n=\text { no of orbit } \\h=\text { plank constant }\end{array}$$
$$\begin{array}{l}(i)=(i i) \\\frac{n h}{2\pi}=\frac{3 q^{2}}{4 \pi \varepsilon_{0} v}\end{array}$$
$$\begin{array}{l}\text { for } n=1 \\\qquad\begin{array}{l}\frac{h}{2 \pi}=\frac{3 q^{2}}{4 \pi \varepsilon_{0} v} \\\text { velocity in } 1^{\text {st }} \text { orbit }(v)=\frac{3 q^{2}}{2 \varepsilon_{0} h}\end{array}\end{array}$$