MCQ Questions for Class 12 Medical Physics Atoms Quiz 13

Let

$v_{1}$ be the frequency of the series limited of the Lyman series,

$v_{2}$ be the frequency of the first line of the Lyman series, and

$v_{3}$ be the frequency of the series limited of the Balmer series. Then:

0%
$v_{1}-v_{2}=v_{3}$
0%
$v_{2}-v_{1}=v_{3}$
0%
$v_{3}=\frac{1}{2} (v_{1}+v_{3})$
0%
$v_{1}+v_{2}=v_{3}$

Explanation

From Rydberg's relation:

$\dfrac{1}{\lambda}=R_H(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

$\implies \nu=R_Hc(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

First line of lyman series is from transition

$2\rightarrow 1$

Series limit of lyman is from transition

$\infty\rightarrow 1$

Series limit of balmer is from transition

$\infty\rightarrow 2$

$\nu_1=R_Hc(1-\dfrac{1}{\infty^2})$

$\nu_2=R_Hc(1-\dfrac{1}{2^2})$

$\nu_3=R_Hc(\dfrac{1}{2^2}-\dfrac{1}{\infty^2})$

$\nu_1-\nu_2=R_Hc(\dfrac{1}{2^2}-\dfrac{1}{\infty^2})=\nu_3$

$A_{n}$ is the area enclosed in the

$n$ th orbit in a hydrogen atom then the graph log

$\left ( \frac{A_{n}}{A_{1}} \right )$ against log

$n$ :

0%
will have slope 2 (a straight line)
0%
will have slope 4 (a straight line)
0%
will be a monotonically increasing non-linear curve
0%
will be a circle

Explanation

Concept attached

$r_n=Cn^2$

$r_1 = C$

$r_n= r_1 n^2$

$A_n= \pi r_n ^2 = \pi r_1 ^2 n^4$

$log(\cfrac{A_n}{A_1}) = 4 logn +K$
Slope =4, Linear graph.

Let

$m_{P}$ be the mass of a proton,

$m_{n}$ the mass of a neutron,

$M_{1}$ the mass of a

$_{10}^{20}Ne$ nucleus and

$M_{2}$ the mass of a

$_{20}^{40} Ca$ nucleus. Then

0%
$M_{2} = 2M_{1}$
0%
$M_{2} > 2M_{1}$
0%
$M_{2} < 2M_{1}$
0%
$M_{1} < 10(m_{n} + m_{p})$

Explanation

$T \propto \dfrac{n}{\dfrac{1}{n^2}}$

$\longrightarrow T \propto n^3$

$\dfrac{T_1}{T_2} = \dfrac{n_1 ^3 } { n_2 ^3 }$

$n_1 ^3 = 8 n_2 ^3$

$n_1 = 2n_2$

From Options , A and D are possible .

The quantum number n of the state finally populated in

$He^{+}$ ions is

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$\Delta E_{h}=\dfrac{3}{4}\times 13.6\ eV =$ Energy released by H atom.

Let He

$^{+}$ go to nth state.

So energy required,

$\Delta E_{he}= 13.6 \times 4\left ( \dfrac{1}{4}-\dfrac{1}{n^{2}} \right )$ eV

$\Delta E_{he}=\Delta E_{h}$

$\Rightarrow \dfrac{3}{4}\times 13.6=13.6\times 4\left ( \dfrac{1}{4}-\dfrac{1}{n^{2}} \right )$

$\Rightarrow$

$n=4$

An electron in a hydrogen atom makes a transition from

$n=n_{1}$ to

$n=n_{2}$ . The time period of the electron in the initial state is eight times that in the final state. The possible values of

$n_{1}$ and

$n_{2}$ are

0%
$n_{1}=4, n_{2}=2$
0%
$n_{1}=8, n_{2}=2$
0%
$n_{1}=8, n_{2}=1$
0%
$n_{1}=6, n_{2}=3$

If elements of quantum number greater than

$'n'$ were not allowed, the number of possible elements in nature would have been

0%
$\displaystyle\frac{1}{2}n(n+1)$
0%
$\left\{\displaystyle\frac{n(n+1)}{2}\right\}^2$
0%
$\displaystyle\frac{1}{6}n(n+1)(2n+1)$
0%
$\displaystyle\frac{1}{3}n(n+1)(2n+1)$

Explanation

${ n }^{ th }$ shell can hold upto

$2{ n }^{ 2 }$ electrons
Total number of possible elements would be

$\sum _{ 1 }^{ n }{ 2{ n }^{ 2 } }$

$=\dfrac { n }{ 3 } \left( n+1 \right) \left( 2n+1 \right)$

Monochromatic radiations of wavelength

$\lambda$ are incident on a hydrogen sample in the ground state. Hydrogen atom absorbs the light and subsequently emits radiations of

$10$ different wavelengths. The value of

$\lambda$ is nearly

0%
$203\space nm$
0%
$95\space nm$
0%
$80\space nm$
0%
$73\space nm$

Explanation

Total no. of wavelengths emitted

$=\ ^n C _2= \cfrac{n(n-1)}{2}$

$^n C _2 = 10$

$n^2 -n -20 =0$

$n=-4 \ or \:5$

$n=5$

So the energy required:

$\cfrac{1}{\lambda} = R \left (1 - \cfrac{1}{5^2}\right )$

$R = 1.1 \times 10^7$

$\lambda = 94.6\ nm$

Check the correctness of the following statements about the Bohr Model of hydrogen atom:

$(i)$ The acceleration of the electron in

$n=2$ orbit is more than that in

$n=1$ orbit.

$(ii)$ The angular momentum of the electron in

$n=2$ orbit is more than that in

$n=1$ orbit.

$(iii)$ The KE of the electron in

$n=2$ orbit is less than that in

$n=1$ orbit

0%
All the statements are correct
0%
Only $(i)$ and $(ii)$ are correct
0%
Only $(ii)$ and $(iii)$ are correct
0%
Only $(iii)$ and $(i)$ are correct

Explanation

Angular momentum,

$L=mvr_n=\dfrac{nh}{2\pi}\propto n$
Hence, (ii) is correct.
Velocity of

$2^{nd}$ orbit is less so kinetic energy is also less.

Hence, (iii) is correct.
The radius of

$2^{nd}$ orbit is

$4$ times the radius of

$1^{st}$ orbit.
Thus, acceleration

$(v^2/r)$ in

$n=2$ is less than the

$n=1$ .
Hence, (i) is incorrect.

Find the value of principal quantum number n corresponding 20 kg satellite.

0%
$1.6\times 10^{45}$
0%
$1.6\times 10^{8}$
0%
$7.6\times 10^{5}$
0%
$8.3\times 10^{11}$

Explanation

We know that:

$nh=mvR$ , where

$R=$

$8060\times 10^3 m$

Also,

$v=\sqrt{\dfrac{GM}{R}}$ ,

$M =$ mass of earth

$=6\times 10^24kg$

$n = \displaystyle \frac{m}{h} \sqrt{\displaystyle \frac{GM}{R}} R = \displaystyle \frac{20}{6.62\times 10^{-34}}\sqrt{6.67\times 10^{-11}\times 6\times 10^{24}\times 6400\times 10^3}\ \ 8060\times 10^3$

$= \displaystyle \frac{20\times 10^{34}}{6.62}\times 10^9\times 8\times \sqrt {40} \\ = 1.6\times 10^{45}$

Find the radius of first two allowed orbits.

0%
$2.5 \times 10^{-84}m, 10^{-83}m$
0%
$400 km, 1600 km$
0%
$200 m, 800 m$
0%
$3 km, 12 km$
0%
none

Explanation

$mvR=nh$

$V_o = \sqrt{\cfrac{GM}{R}}$
Substitute:

$m\sqrt{\dfrac{GM}{R}}R=nh\\ R = \cfrac{n^2h^2}{m^2GM}$

$R = n^2 k$

$k = 1.6 \times 10^{-84}$ m

$R_1$ = 1.6

$\times 10^{-84} m$

$R_2 = 6.4 \times 10^{-84}$ m

If potential energy between a proton and an electron is given by

$|U| = ke^2/2R^3$ , where

$e$ is the charge of electron and

$R$ is the radius of atom, then radius of Bohr's orbit is given by (

$h$ = Planck's constant,

$k$ = constant)

0%
$\displaystyle\frac{ke^2m}{h^2}$
0%
$\displaystyle\frac{6\pi^2}{n^2} \displaystyle\frac{ke^2m}{h^2}$
0%
$\displaystyle\frac{2\pi}{n} \displaystyle\frac{ke^2m}{h^2}$
0%
$\displaystyle\frac{4\pi^2ke^2m}{n^2h^2}$

Explanation

Potential energy at a distance

$R$ is given as,

$U=\dfrac{ke^2}{2R^3}$

Thus force between proton and electron

$=-\dfrac{dU}{dR}$

$=\dfrac{3ke^2}{2R^4}$

This force causes the centrifugal acceleration of the system.

Thus,

$\dfrac{3ke^2}{2R^4}=\dfrac{mv^2}{R}$

Bohr's postulate is:

$mvR=\dfrac{nh}{2\pi}$

Substituting the expression for

$v$ from above equation gives,

$\dfrac{3ke^2}{2R^4}=\dfrac{m}{R}(\dfrac{nh}{2\pi m R})^2$

$\implies R=\dfrac{6ke^2\pi^2m}{n^2h^2}$

Which ray/radiation can be detected by GM counter?

0%
Alpha
0%
Beta
0%
$\gamma$ ray
0%
all

The wavelength of

$K_\alpha$ X-rays of two metals

$A$ and

$B$ are

$\dfrac{4}{1875R}$ and

$\dfrac{1}{675R}$ , respectively, where

$\mbox{'R'}$ is Rydberg's constant. The number of elements lying between

$A$ and

$B$ according to their atomic number is

0%
$3$
0%
$6$
0%
$5$
0%
$4$

Explanation

In X-ray spectroscopy, K-alpha emission lines result when an electron transitions to the innermost "K" shell (principal quantum number 1) from a 2p orbital of the second or "L" shell (with principal quantum number 2).

$\cfrac{1}{\lambda}= Z^2 R(1 - \cfrac{1}{2^2}) = \cfrac{3RZ^2}{4}$

$\lambda _A = \cfrac{4}{1875R} = \cfrac{4}{3 R Z^2}$

$Z_A =25$

Similarly,

$\lambda _B = \cfrac{1}{675R} = \cfrac{4}{3 R Z_B ^2}$

$Z_B = 30$

No. of elements between A and B

$= 4$

In a hydrogen atom, electron is in the

$n^{th}$ excited state. It comes down to the first excited state by emitting

$10$ different wavelengths. The value of

$n^{th}$ is

0%
$6$
0%
$7$
0%
$8$
0%
$9$

Explanation

Total no. of levels

$= n-1$

$n \rightarrow 2$
Put

$n' = n-1$ in the formula attached

$\cfrac{(n-1) (n-2)}{2} = 10$

$(n-1)(n-2) = 20$

$\Rightarrow n=6$

Which of the following statement about hydrogen spectrum are correct?

0%
All the lines of Lyman series lie in ultraviolet region
0%
All the lines of Balmer series lie in visible region
0%
All the lines of Paschen series lie in infrared region
0%
None of the above

Explanation

As per the data shown in the figure, and the wavelenghts that can be determined using the Rydberg formula, the correct ans. can be obtained

$\cfrac{1}{\lambda} = Z^2 R (\cfrac{1}{n_1 ^2}- \cfrac{1}{n_2 ^2})$

If, in a hydrogen atom, radius of

$n^{th}$ Bohr orbit is

$r_n$ , frequency of revolution of electron in

$n^{th}$ orbit if

$f_n$ , and area enclosed by the

$nth$ orbit is

$A_n$ , then which of the following graphs are correct?

Explanation

Radius

$r = cn^2$

Area
Area =

$A(n) = \pi r^2 = C \pi n^4$

$A_1 = C \pi$

$y = ln (A_n / A_10) = ln(n^4) = 4 ln(n)$

$x = ln(n)$
Thus the equation becomes:

$y=4x$

Frequency

$v \propto \cfrac{1}{n^3}$

$f(n) = \cfrac{C}{n^3}$

$f(1) = C$

$y =ln(\cfrac{f(n)}{f(1)} = ln(1/n^3)= -3ln(n)$

$y = -3x$
Negative slope.

According to Bohr's theory of hydrogen-atom, for the electron in the

$n^{th}$ permissible orbit

0%
$\mbox{Linear Momentum } \propto \displaystyle\frac{1}{n}$
0%
$\mbox{Radius of orbit } \propto n$
0%
$\mbox{Kinetic Energy } \propto \displaystyle\frac{1}{n^2}$
0%
$\mbox{Angular Momentum } \propto n$

Explanation

Angular Momentum Quantisation:

$mvr = \cfrac{nh}{2 \pi}$ ------------(1)

Centripetal force:

$\cfrac{k(e)e}{r^2} = \cfrac{mv^2}{r}$ ----------------(2)

$v^2 r = \cfrac{ke^2}{m}$

Substituting (1) in (2)

$v= \cfrac{2 \pi ke^2}{n h}$

Energy:

$E = 13.6 \cfrac{Z^2}{n^2}$ -----------(3)

Linear Momentum:

$mvr = nh/2 \pi$

$mv = \cfrac{nh}{2 \pi r}$ ---------------(4)

From the concept attached:

$r \propto n^2$ -----------------(5)

Hence,

$mv \propto \cfrac{1}{n}$ by (6)
and

$U = 2KE$

$E \propto \cfrac{1}{n^2}$ (concept attached)

Which of the following products, in a hydrogen atom, are independent of the principal quantum number

$n$ ? The symbols have their usual meanings.

0%
$\omega^2r$
0%
$\displaystyle\frac{E}{v^2}$
0%
$v^2r$
0%
$\displaystyle\frac{E}{r}$

Explanation

Angular Momentum Quantisation:

$mvr = \cfrac{nh}{2 \pi}$ ------------(1)

Centripetal force:

$\cfrac{k(e)e}{r^2} = \cfrac{mv^2}{r}$ ----------------(2)

$v^2 r = \cfrac{ke^2}{m}$ (Constant)

Substituting (1) in (2):

$v= \cfrac{2 \pi ke^2}{n h}$

$v = \cfrac{C_1}{n}$

Energy:

$E = 13.6 \cfrac{Z^2}{n^2}$ -----------(3)

$\cfrac{E}{v^2}= const.$

Let

$A_n$ be the area enclosed by the

$n^{th}$ orbit in a hydrogen atom. The graph of

$ln(A_n/A_l)$ against

$ln(n)$

0%
Will pass through origin
0%
Will be a straight line with slope $4$
0%
Will be monotonically increasing nonlinear curve
0%
Will be circle

Explanation

For hydrogen atom , radius

$r_n \propto n^2$

So,

$\dfrac{A_n}{A_1}=\dfrac{\pi r_n^2}{\pi r_1^2}=n^4/1^4=n^4$

$ln(A_n/A_1)=ln(n)^4=4ln(n)$

This is straight line equation

$(y=mx)$ with slope m=4 and it will pass through origin because it has no intercept.

Mark out the correct statement(s).

0%
Line spectra contain information about atoms only
0%
Line spectra contain information about both atoms and molecules
0%
Band spectra contain information about both atoms and molecules
0%
Band spectra contain information about molecules only

Continuous spectrum is produced by

0%
Incandescent electric bulb
0%
Sun
0%
Hydrogen molecules
0%
Sodium vapor lamp

In Bohr's model of the hydrogen atom, let

$R,V,T,$ and

$E$ represent the radius of the orbit, speed of the electron, time period of the revolution of electron and the total energy of the electron, respectively. The quantities proportional to the quantum number

$n$ are

0%
$VR$
0%
$RE$
0%
$\displaystyle\frac{T}{R}$
0%
$\displaystyle\frac{V}{E}$

Explanation

Radius of n th orbit :

$R \propto \dfrac{n^2}{Z}= n^2 ...(1)$ (for hydrogen atom

$Z=1$ )

Velocity,

$V \propto Z/n= 1/n ...(2)$ ;

Time period

$T=\dfrac{2\pi R}{V}$

Using (1) and (2),

$T\propto n^3 ...(3)$

Total energy ,

$E \propto Z^2/n^2=1/n^2 .....(4)$

Using (1) and (2)

$VR \propto n$

Using (1) and (4)

$RE=1$

Using (1) and (3)

$T/R \propto n$

Using (2) and (4)

$V/E \propto n$

Let the energy of the electron confined in the atom be

$E$ . At what

$F_0$ would the atom ionize?

0%
$\displaystyle F_0 = \frac {E^2 \pi \varepsilon_0} {e}$
0%
$\displaystyle F_0 = \frac {E^2 \varepsilon_0} {e^3}$
0%
$\displaystyle F_0 = \frac { \pi \varepsilon_0} {e^3}$
0%
$\displaystyle F_0 = \frac {E^2 \pi \varepsilon_0} {e^3}$

If the atom

$_{100} Fm^{257}$ follows the Bohr model and the radius of

$_{100} Fm^{257}$ is n times the Bohr radius, then find

$n$ .

0%
$100$
0%
$200$
0%
$4$
0%
$1/4$

Explanation

For an atom following Bohr's model, the radius is given by

$\displaystyle r_m=\frac{r_0m^2}{Z}$

where

$r_0=$ Bohr's radius and

$m=$ orbit number.
For

$Fm, m=5$ (Fifth orbit in which the outermost electron is present)

$\displaystyle\therefore r_m=\frac{r_05^2}{100}=nr_0$ (given)

$\displaystyle\Rightarrow n=\frac{1}{4}$

The orbital radius of the first excited level of positronium atom is

Note:

$a_0$ is the orbital radius of ground state of hydrogen atom

0%
$4a_0$
0%
$a_0/2$
0%
$8a_0$
0%
$2a_0$

Explanation

Given, electron in the positronium has reduced mass

$\dfrac { { m }_{ e } }{ 2 }$
And orbital radius of ground state of hydrogen atom is

${ a }_{ 0 }$ .
Therefore, for positronium,

${ r }_{ n }=2{ a }_{ 0 }{ n }^{ 2 }$
Hence, orbital radius of first excited level of positronium is

$8{ a }_{ 0 }$ .

What principle is violated here?

0%
Laws of Motion
0%
Energy conservation
0%
Nothing is violated
0%
Cannot be decided

Explanation

Laws of Motion:
Newtons second law:

$F = ma$
Without any force, the atom cannot accelerate.
Conservation of Momentum:
The linear momentum of the atom can't change without any external force acting on it.

The ratio between Bohr radii is

0%
1 : 2 : 3
0%
2 : 4 : 6
0%
1 : 4 : 9
0%
1 : 3 : 5

Explanation

Radius

$\propto n^2$
Hence,

$R_1:R_2:R_3 = 1:4:9$

The radius of the first Bohr orbit is

0%
$0.529\times10^{-10}\space m$
0%
$0.106\times10^{-10}\space m$
0%
$0.318\times10^{-10}\space m$
0%
None of these

Explanation

Given,

$47.2\ eV$ is required to excite an electron from

$n=2$ to

$n=3$

$\Rightarrow \dfrac { 13.6 { Z }^{ 2 } }{ 4 } -\dfrac { 13.6 { Z }^{ 2 } }{ 9 } =\dfrac { 5\times 13.6 { Z }^{ 2 } }{ 36 } =47.2$

$\therefore \ { Z }^{ 2 }= 24.99$

Therefore,

$Z=5$

Radius of first Bohr orbit is

$=0.529\times \dfrac { { 1 }^{ 2 } }{ 5 } =0.1058\mathring { A }$

$\Rightarrow \ { r }_{ 1 }= 0.1058\times { 10 }^{ -10 }m$

In which transition is a Balmer series photon absorbed?

0%
$II$
0%
$III$
0%
$IV$
0%
$VI$

Explanation

Absorption occurs when electron excites to higher level.
For Balmer, electron has to excite from

$n=2$ .

The minimum permissible radius of the orbit will be :

0%
$\displaystyle\frac{2\epsilon_0h^2}{m\pi e^2}$
0%
$\displaystyle\frac{4\epsilon_0h^2}{m\pi e^2}$
0%
$\displaystyle\frac{\epsilon_0h^2}{m\pi e^2}$
0%
$\displaystyle\frac{\epsilon_0h^2}{2m\pi e^2}$

Explanation

Minimum radius will be when

$n=1$ :

$mvR_\circ{}=2 \dfrac{h}{2\pi}\Rightarrow vR_\circ{}= \dfrac{h}{m\pi}$ ...(i)

Electrostatic force is providing centripetal force for electron's circular motion:

$\dfrac{1}{4\pi\epsilon_\circ{}}\dfrac{e^2}{R_\circ{}^2}=\dfrac{mv^2}{R_\circ{}}$ ...(ii)

$\dfrac{1}{4\pi\epsilon_\circ{}}\dfrac{e^2}{m}=v^2R_\circ{}$ ...(iii)

Substituting

$vR_\circ{}$ from equation (i)

$v \dfrac{h}{m\pi}=\dfrac{1}{4\pi\epsilon_\circ{}}\dfrac{e^2}{m}$

$v =\dfrac{e^2}{4\epsilon_\circ{}h}$

From relation (i)

$R_\circ{}=\dfrac{h}{m\pi v}$

$R_\circ{}=\dfrac{h^2 4\epsilon_\circ{}}{me^2\pi}$

Two smooth tunnels are dug from one side of the earths surface to the other side, one along a diameter and the other along a chord. Now two particles are dropped from one end of each of the tunnels. Both the particles oscillate simple harmonically along the tunnels. Let

${T}{1}$ and

${T}{2}$ be the time periods and

${v}_{1}$ and

${v}_{2}$ be the maximum speed of the particles in the two tunnels. Then

0%
${T}_{1}={T}_{2}$
0%
${T}_{1}>{T}_{2}$
0%
${v}_{1}={v}_{2}$
0%
${v}_{1}>{v}_{2}$

A donor atom in a semiconductor has a loosely bound electron. The orbit of this electron is considerably affected by the semiconductor material but behaves in many ways like an electron orbiting a hydrogen nucleus. Given that the electrons has an effective mass of

$0.07\ m_e$ (where

$m_{e}$ is mass of the free electron) and the space in which it moves has a permittivity

$13\epsilon_{0}$ , then the radius of the electron's lowermost energy orbit will be close to (The Bohr radius of the hydrogen atom is

$0.53\overset {\circ}{A})$ .

0%
$0.53\overset {\circ}{A}$
0%
$243\overset {\circ}{A}$
0%
$10\overset {\circ}{A}$
0%
$100\overset {\circ}{A}$

Explanation

$\dfrac {mv^{2}}{r} = \dfrac {KQ^{2}}{r^{2}}$
and

$mvr = \dfrac {nh}{2\pi}\implies mv = \dfrac{nh}{2 \pi r}$

and

$\dfrac{m^2v^2}{mr}=\dfrac{KQ^2}{r^2}$

$\dfrac{n^2h^2}{4 \pi^2mr^3}=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q^2}{r^2}$

$\dfrac{n^2h^2}{4\pi^2(0.07m_e)r}=\dfrac{1}{4\pi(13) \epsilon_0} Q^2$

$r=\dfrac{13n^2h^2}{4\pi^2kQ^2(0.07)m_e}$

$r=\dfrac{13}{0.07} \times r_0$ where

$r_0=bohr \ radius$

$r=\dfrac{13}{0.07} \times 0.53 \overset{\circ}{A}$

$r \approx 100\overset {\circ}{A}$ .

The radius of the first orbit of hydrogen is

$r_H$ , and the energy in the ground state is

$-13.6$ eV. Considering a

$\mu^-$ -particle with a mass

$207$

$m_e$ revolving round a proton as in Hydrogen atom, the energy and radius of proton and

$\mu^-$ combination respectively in the first orbit are:

[Assume nucleus to be stationary]

0%
$-13.6\times 207\ eV$ , $\displaystyle\frac{r_H}{207}$
0%
$-207\times 13.6\ eV$ , $207$ $r_H$
0%
$\displaystyle\frac{-13.6}{207}\ eV$ , $\displaystyle\frac{r_H}{207}$
0%
$\displaystyle\frac{-13.6}{207}\ eV$ , $207$ $r_H$

Explanation

Radius of ground state,

$r = \dfrac{h^2}{4\pi ^2 mKe^2}$

$\implies$

$r \propto \dfrac{1}{m}$

$\implies$

$\dfrac{r_{\mu}}{r_H} = \dfrac{m_e}{m_{\mu}} = \dfrac{m_e}{207 m_e}$

$\implies$

$r_{\mu } = \dfrac{r_H}{207}$

Energy of ground state,

$E = - \dfrac{Ke^2 }{2 r}$

$\implies$

$E \propto m$

$(\because r \propto \dfrac{1}{m})$

$\implies$

$\dfrac{E_{\mu}}{ -13.6} = \dfrac{m_{\mu}}{m_e} = \dfrac{207m_e}{ m_e}$

$\implies$

$E_{\mu} = -13.6 \times 207$

$eV$

In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If

$a_0$ is the radius of the ground state orbit, m is the mass and e is the charge on the election and

$\varepsilon_0$ is the vacuum permittivity, the speed of the electron is?

0%
$0$
0%
$\displaystyle\frac{e}{\sqrt{\varepsilon_0a_0m}}$
0%
$\displaystyle\frac{e}{\sqrt{4\pi\varepsilon_0a_0m}}$
0%
$\sqrt{\displaystyle\frac{4\pi\varepsilon_0a_0m}{e}}$

Explanation

Centripetal force

$=\dfrac { m{ v }^{ 2 } }{ { a }_{ \circ } } \rightarrow (1)$

Colounbic force

$=\dfrac { 1 }{ 4\pi { \varepsilon }_{ \circ } } \dfrac { e.e }{ { a }^{ 2 } } \rightarrow (2)$

$\therefore$ Centripetal force= Coloumbic force

$\rightarrow \dfrac { m{ v }^{ 2 } }{ { a }_{ \circ } } \dfrac { 1 }{ 4\pi { \varepsilon }_{ \circ } } \dfrac { { e }^{ 2 } }{ { { a }_{ \circ } }^{ 2 } } \\ \rightarrow { v }^{ 2 }=\dfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ \circ }{ a }_{ \circ }m } \\ or\quad v=\dfrac { e }{ \sqrt { 4\pi { \varepsilon }_{ \circ }{ a }_{ \circ }m } }$

To calculate the size of a hydrogen anion using the Bohr model, we assume that its two electrons move in an orbit such that they are always on diametrically opposite sides of the nucleus. With each electron having the angular momentum

$\hbar = h/2\pi$ , and taking electron interaction into account the radius of the orbit in terms of the Bohr radius of hydrogen atom

$a_{B}= \dfrac {4\pi \epsilon_{0}\hbar^{2}}{me^{2}}$ is

0%
$a_{B}$
0%
$\dfrac {4}{3}a_{B}$
0%
$\dfrac {2}{3}a_{B}$
0%
$\dfrac {3}{2}a_{B}$

Explanation

Let the velocity of the electron be

$v$ .
Angular momentum of electron

$mvr = \hbar$ (given)
Net electrostatic force acting on the electron

$F_e = \dfrac{e^2}{4\pi\epsilon_o r^2} - \dfrac{e^2}{4\pi\epsilon_o (2r^2)} = \dfrac{3}{4}\dfrac{e^2}{4\pi\epsilon_o r^2}$
Centrifugal force

$F_c = \dfrac{mv^2}{r} = \dfrac{(mvr)^2}{r\times mr^2} = \dfrac{\hbar^2}{mr^3}$
But

$F_c = F_e$

$\therefore$

$\dfrac{\hbar^2}{mr^3} = \dfrac{3}{4}\dfrac{e^2}{4\pi\epsilon_o r^2}$

$\implies \ r = \dfrac{4}{3}\dfrac{4\pi\epsilon_o \hbar^2}{me^2} = \dfrac{4}{3}a_B$

The number of revolutions per second made by an electron in the first Bohr's orbit of hydrogen atom is(Given,

$r=0.53\overset{o}{A}$ ).

0%
$6\times 10^5$ Hz
0%
$2.5\times 10^{12}$ Hz
0%
$1.9\times 10^{13}$ Hz
0%
$6.5\times 10^{15}$ Hz

Explanation

From Bohr's theory

$mvr=\displaystyle\frac{nh}{2\pi}$
For first orbit

$n=1$

$m(r\omega )r=\displaystyle\frac{1\times h}{2\pi}$

$\Rightarrow mr^22\pi f=\displaystyle\frac{h}{2\pi}$

$\Rightarrow \displaystyle f=\frac{h}{4\pi^2mv^2}$

$=\displaystyle\frac{6.6\times 10^{-34}}{4\times (3.14)^2\times 9.1\times 10^{-31}\times (0.33\times 10^{-10})^2}$ .

$6.5\times10^{15} Hz$

Which of the following statement is wrong?

0%
$\Psi_s$ = - $\pi$
0%
DPD = -OP + TP
0%
$\Psi_w$ = $\Psi_s$ + $\Psi_p$
0%
$\Psi_w$ = -DPD

$\dfrac{h}{ 2 \pi}$ , and taking electron interaction into account the radius of the orbit in terms of the Bohr radius of hydrogen atom

$a_B=\displaystyle\frac{4\pi \epsilon_0h^2}{me^2}$ is?

0%
$a_B$
0%
$\displaystyle\frac{4}{3}a_B$
0%
$\displaystyle\frac{2}{3}a_B$
0%
$\displaystyle\frac{3}{2}a_B$

If velocity of

$\alpha -$ particle is made

$1/3$ times, then % variation in distance of closest approach will be: (with respect to initial)

0%
300%
0%
600%
0%
800%
0%
80%

Explanation

The distance of closest approach of a particle is given by,

$r=\dfrac { 4KZ{ e }^{ 2 } }{ m{ v }^{ 2 } }$

Now, in second case, when

$v\rightarrow \dfrac { v }{ 3 }$ , distance of closed approach will be,

${ r }^{ ' }=\dfrac { 4kZ{ e }^{ 2 } }{ m{ \left( \dfrac { v }{ 3 } \right) }^{ 2 } } =\dfrac { 4kZ{ e }^{ 2 } }{ m{ { v } }^{ 2 } } =9r$

If r represents

$100$ %

then

$9r=900$ %.

Percentage variation

$=900-100=800$ %

The radius of the smaller electron orbit in hydrogen-like ion is

$\dfrac{0.51\times 10^{-10}}{4}$ m, then it is?

0%
hydrogen atom
0%
$He^+$
0%
$Li^+$
0%
$Be^{3+}$

Explanation

For hydrogen like atom, the radius of nth orbit

$r^z_n=\displaystyle\frac{n^2}{Z}a_0$
Here,

$a_0=0.51\times 10^{-10}$ m

$\therefore r^z_n=\displaystyle\frac{0.51\times 10^{-10}}{4}$ m
In the ground state,

$n=1$

$\therefore \displaystyle\frac{0.51\times 10^{-10}}{4}=\frac{1^2}{z}\times 0.51\times 10^{-10}$

$\therefore Z=4$
So, the atom is triply ionised beryllium

$(Be^{3+})$ .

Given mass number of glod=197, density of gold=19.7 g per

$cm^3$ , Avogadro's number =

$6\times 10^{23}$ .The radius of the gold atom is approximately:

0%
$1.5 \times 10^{-8}m$
0%
$1.5 \times 10^{-9}m$
0%
$1.5 \times 10^{-10}m$
0%
$1.5 \times 10^{-12}m$

In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as

$\quad F=\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 } } \left( \cfrac { 1 }{ { r }^{ 2 } } +\cfrac { \beta }{ { r }^{ 3 } } \right)$ , where

$\beta$ is a constant. For this atom, the radius of the

$n$ th orbit in terms of the Bohr radius

$\left( { a }_{ 0 }=\cfrac { { \varepsilon }_{ 0 }{ h }^{ 2 } }{ m\pi { e }^{ 2 } } \right)$ is:

0%
${ r }_{ n }={ a }_{ 0 }n-\beta$
0%
${ r }_{ n }={ a }_{ 0 }{ n }^{ 2 }+\beta \quad$
0%
${ r }_{ n }={ a }_{ 0 }{ n }^{ 2 }-\beta$
0%
${ r }_{ n }={ a }_{ 0 }n+\beta$

Explanation

$\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 } } \left( \cfrac { 1 }{ { r }^{ 3 } } -\cfrac { \beta }{ { r }^{ 3 } } \right)$

$\quad \therefore { v }^{ 2 }=\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }mr } \left( \cfrac { 1 }{ { r }^{ } } +\cfrac { \beta }{ { r }^{ 2 } } \right)$
From Bohr's hypothesis

$\quad v=\cfrac { nh }{ 2\pi mr } \quad \therefore \cfrac { { n }^{ 2 }{ hn }^{ 2 } }{ 4{ \pi }^{ 2 }{ m }^{ 2 }{ r }^{ 2 } } =\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }mr } \left( \cfrac { 1 }{ { r }^{ } } +\cfrac { \beta }{ { r }^{ 2 } } \right)$

$\therefore r+\beta ={ a }_{ 0 }{ n }^{ 2 }\quad$

A diatomic molecules has moment of inertia

$I$ . By Bohr's quantization condition its rotational energy in the

$n^{th}$ level

$(n = 0$ is not allowed) is

0%
$\dfrac {1}{n^{2}}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$
0%
$\dfrac {1}{n}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$
0%
$n\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$
0%
$n^{2}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the

$n$ th orbital will be proportional to

0%
${ n }^{ 2 }$
0%
$n$
0%
${ n }^{ 1/2 }$
0%
${ n }^{ 1/4 }$

Explanation

$qvB=\cfrac { m{ v }^{ 2 } }{ r } ;mvr=\cfrac { nh }{ 2\pi }$

$\therefore \cfrac { qBr }{ m } =\cfrac { nh }{ 2\pi mr }$

$\therefore r\propto { n }^{ 1/2 }\quad$

An electron in a hydrogen atom makes a transition from

$n=n_{1}$ to

$n=n_{2}$ . The time period of the electron in the initial state is eight times that in the final state. The possible values of

$n_{1}$ and

$n_{2}$ are

0%
$n_{1}=4, n_{2}=2$
0%
$n_{1}=8, n_{2}=2$
0%
$n_{1}=8, n_{2}=1$
0%
$n_{1}=6, n_{2}=3$

Explanation

$\textbf{Correct option : option A and D}$

$\textbf{Explanation:}$

We know $T\propto n^3$ thus ,

$\dfrac{T_1}{T_2} \propto (\dfrac{n_1}{n_2})^3$

$\dfrac{8}{1} \propto (\dfrac{n_1}{n_2})^3$

(because here $\dfrac{T_1}{T_2}$ is $\dfrac{8}{1}$ )

$\Rightarrow \dfrac{n_1}{n_2} = \dfrac{2}{1}$

thus the options in which $n_1=2n_2$ are the correct options.

$\textbf{Thus Option A and option D is correct .}$

A mixture of

$2$ moles of helium gas (atomic mass

$=4$ amu) and

$1$ mole of argon gas (atomic mass

$=40$ amu) is kept at

$300$ K in a container. The ratio of the rms speeds

$\left(\dfrac{v_{rms}(helium)}{v_{rms}(argon)}\right)$ is?

0%
$0.32$
0%
$0.45$
0%
$2.24$
0%
$3.16$

Explanation

$V_{rms}=\sqrt{\cfrac{3RT}{M}}$

$V_{rms}$ for helium $=\sqrt{\cfrac{3RT}{4}}$

$V_{rms}$ for argon $=\sqrt{\cfrac{3RT}{40}}$

$\cfrac{V_{rms}(helium)}{V_{rms}(Argon)}=\sqrt{10}$

$=3.16$

The wavelengths involved in the spectrum of deuterium

$\left( _{ 1 }^{ 2 }{ D } \right)$ are slightly different from that of hydrogen spectrum, because

0%
size of the two nuclei are different
0%
nuclear forces are different in the two cases
0%
masses of the two nuclei are different
0%
attraction between the electron and the nucleus is different in the two cases

In certain electronic transition from quantum level n to ground state in atomic hydrogen in one or more steps no line belonging to Brackett series is observed. The wave numbers which may be observed in Balmer series is then?

0%
$\dfrac{8R}{9}, \dfrac{5R}{36}$
0%
$\dfrac{3R}{16}, \dfrac{8R}{9}$
0%
$\dfrac{5R}{36}, \dfrac{3R}{16}$
0%
$\dfrac{3R}{4}, \dfrac{3R}{16}$

Explanation

$Given:$ In certain electronic transition from quantum level n to ground state in atomic hydrogen in one or more steps no line belonging to Brackett series is observed.

$Solution:$

The

$n$ should be equal to 4 ; otherwise if

$n>4$ ,

then Bracket series will also be noticed.

Total no.of lines formed

$=\mathbf{n}(\mathbf{n}-1) / 2=4 \times 3 / 2=6$

No. of lines formed will be due to transition

between these energy levels

$\mathbf{n}=4$ to

$\mathbf{n}=\mathbf{3}$

$\mathbf{n}=4 \operatorname{ton}=2$

$\mathbf{n}=4$ to

$\mathbf{n}=\mathbf{1}$

$\mathbf{n}=3 \operatorname{ton}=2$

$\mathbf{n}=3 \operatorname{ton}=1$

$\mathbf{n}=2$ to

$\mathbf{n}=1$

Balmer series line obtained when any electron comes to second orbit from outer orbit so

out of these six lines, only two lines belongs to Balmer series.

i.e., from

$n=4$ to

$n=2$ and

$\mathbf{n}=3$ to

$\mathbf{n}=2$

For these lines wave number is,

$(\mathrm{i}) \overline{\mathbf{v}{1}}=\frac{\mathbf{1}}{\lambda}=\mathbf{R}{\mathbf{H}}\left[\frac{\mathbf{1}}{\mathbf{2}^{2}}-\frac{\mathbf{1}}{\mathbf{4}^{2}}\right]=\frac{\mathbf{1 2}}{\mathbf{6 4}} \mathbf{R}{\mathbf{H}}=\frac{\mathbf{3}}{\mathbf{1 6}} \mathbf{R}{\mathbf{H}}$

$(\mathrm{ii}) \overline{\mathrm{v}{2}}=\frac{1}{\lambda} \mathbf{R}{\mathrm{H}}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5}{36} \mathbf{R}_{\mathrm{H}}$

$So,the$

$correct$

$option:C$

Calculate the radius of the nth orbit

0%
$\sqrt{\dfrac{nh}{2\pi e B}}$
0%
$\sqrt{\dfrac{nheB}{2\pi}}$
0%
$\sqrt{\dfrac{nhe}{2\pi B}}$
0%
$\sqrt{\dfrac{nhB}{2\pi e}}$

Explanation

According to Bohr's quantization rule

$mvr=\dfrac{nh}{2\pi}$

'r' is less when 'n' has least value i.e. 1

$mv=\dfrac{nh}{2\pi r}.....(1)$

Again,

$r=\dfrac{mv}{qB}$

$mv=rqB.....(2)$

From (1) and (2)

$rqB=\dfrac{nh}{2\pi r}$ [q=e]

$\Rightarrow r^2=\dfrac{nh}{2\pi eB}\Rightarrow r=\sqrt{\dfrac{h}{2\pi eB}}$ [Here, n=1]

For the radius of nth orbit,

$r=\sqrt{\dfrac{nh}{2\pi eB}}$

Angular momentum in second Bohr orbit of H-atom is

$x$ . Then find out angular momentum in 1st excited state of

${Li}^{+2}$ ion:

0%
$3x$
0%
$9x$
0%
$\cfrac{x}{2}$
0%
$x$

Explanation

$\begin{array}{l} \text { we know, } \\ \text { Angular momentum }=\frac{n h}{2 \pi} \end{array}$

$\begin{array}{l} \text { Given, } \\ \text { For } n=2 \\ \text { Angular momentum = } x \\ \qquad x=\frac{2 h}{2 \pi} --(i) \end{array}$

$\text { for } 1^{\text {st }} \text { excited state means } n=2$

$\begin{aligned} & A \cdot M=\frac{2 h}{2 \pi} \quad-(11) \\ \therefore \quad &(i)=(i i) \end{aligned}$

$\text { Angular mementum }= x$

CBSE Questions for Class 12 Medical Physics Atoms Quiz 13 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 13 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.