The relation between $${\lambda}_{1}$$: wavelength of series limit of Lyman series, $${\lambda}_{0}$$: the wavelength of the series limit of Balmer series and $${\lambda}_{3}$$: the wavelength of first line of Lyman series is:

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

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

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

In a $$\mathrm{C}\mathrm{O}$$ molecule, the distance between $$\mathrm{C}$$ (mass $$=12$$ a.m.u) and $$\mathrm{O}$$ (mass $$=16$$ a.m.u.), where 1 a.m.u $$=\displaystyle \frac{5}{3}\times 10^{-27} kg,$$ is close to

$$2.4\times 10^{-10}\mathrm{m}$$

$$1.9\times 10^{-10}\mathrm{m}$$

$$4.4\times 10^{-11}\mathrm{m}$$

The electric potential between a proton and an electron is given by $$V=\displaystyle \mathrm{V}_{0}\ln\frac{\mathrm{r}}{\mathrm{r}_{0}}$$ , where $$\mathrm{r}_{0}$$ is a constant. Assuming Bohr's model to be applicable, write variation of $$\mathrm{r}_{\mathrm{n}}$$ with $$\mathrm{n},\ \mathrm{n}$$ being the principal quantum number?

$$\mathrm{r}_{\mathrm{n}}\propto \mathrm{n}$$

$$\mathrm{r}_{\mathrm{n}}\propto 1/\mathrm{n}$$

$$\mathrm{r}_{\mathrm{n}}\propto \mathrm{n}^{2}$$

$$\mathrm{r}_{\mathrm{n}}\propto 1/\mathrm{n}^{2}$$

MCQ Questions for Class 12 Medical Physics Atoms Quiz 1

According to DE Broglie, wavelength of electron in second orbit is 10

^{- 9}

$^{ -9 }$ meter. Then the circumstances of orbit is :-

0%
10 $^{ -9 }$ m
0%
2×10 $^{ -9 }$ m
0%
3×10 $^{ -9 }$ m
0%
4×10 $^{ -9 }$ m

In a H-atom, the transition takes place from L to K shell. If

R = 1.08 \times 10^{7} m^{- 1}

$R=1.08\times {10}^{7}{m}^{-1}$ , the wave length of the light emitted is nearly

0%
$4400\mathring{A}$
0%
$1250\mathring{A}$
0%
$1650\mathring{A}$
0%
$1850\mathring{A}$

Which of the following particles cannot be deflected by magnetic field ?

0%
Electrons
0%
Neutrons
0%
$\alpha-$ particles
0%
Protons

Explanation

Neutrons do not have charges

\vec{F}

$\vec{F}$ = q

\vec{v}

$\vec{v}$ x

\vec{B}

$\vec{B}$
where q = 0

In which of the following fields cathode rays show minimum deflection ?

0%
Electric field
0%
Magnetic field
0%
Plasma field
0%
Gravitational field

How will you relate velocity of cathode rays to c, if ‘c' denotes the velocity of light?

0%
Equal to c
0%
Greater than c
0%
Less than c
0%
Either greater or less than c

Anti-particle of proton is

0%
Electron
0%
Antiproton
0%
Positron
0%
Neutron

If the K.E. of a cathode ray beam is 8KeV, then the tube should work at a Potential Difference of

0%
4 KV
0%
8 KV
0%
8 V
0%
4 V

Explanation

K.E. = PDx e

\to 8 k e V = P D * e

$\rightarrow 8keV = PD * e$

\to 8 k V = P D

$\rightarrow 8kV = PD$
So, p.d. = 8kV

On decreasing principal quantum number

n

$n$ , the value of

r

$r$ will :

0%
decrease
0%
increase
0%
remain the same
0%
none of the above

Explanation

r = \frac{E_{0} h^{2} n^{2}}{R {m Z e}^{2}}

$r= \dfrac { { E }_{ 0 }{ h }^{ 2 }{ n }^{ 2 } }{ R{ mZe }^{ 2 } }$

So, as

n

$n$ decreases,

r

$r$ will also decrease.

Who proposed the planetary model of the atom?

0%
Earnest Rutherford
0%
Max Planck
0%
Neils Bohr
0%
James Chadwick
0%
J.J. Thompson

In Rutherford experiment, most of the alpha particles go straight through the foil because________

0%
Alpha particles are much heavier than electron.
0%
Alpha particles are positively charged.
0%
Alpha particles move with high velocity.
0%
Most part of the atom is empty.

In terms of Bohr radius

a_{0}

${a}_{0}$ , the radius of the second Bohr orbit of a hydrogen atom is given by:

0%
$8{a}_{0}$
0%
$4{a}_{0}$
0%
$2{a}_{0}$
0%
$\sqrt {2}{a}_{0}$

Explanation

Since,

r \propto n^{2}

$r \propto {n^2}$

Radius of

2^{n d}

$2^{nd}$ Bohr's orbit

= 4 a_{0}

$= 4a_0$ .

The photon radiated from hydrogen corresponding to

2 n d

$2nd$ line of Lyman series is absorbed by a hydrogen like atom

^{'} X^{'}

$'X'$ in

2 n d

$2nd$ excited state. As a result the hydrogen like atom

^{'} X^{'}

$'X'$ makes a transition to

n^{t h}

$n^{th}$ orbit. Then,

0%
$X=He^+, n=4$
0%
$X=Li^{++}, n=6$
0%
$X=He^+, n=6$
0%
$X=Li^{++}, n=9$

The hydrogen atom in ground state is excited by a monochromatic radiation of

λ = 975 A^{\circ}

$\lambda=975 A^{\circ}$ . Number of spectral lines in the resulting spectrum emitted will be

0%
3
0%
2
0%
6
0%
10

An excited electron of H-atoms emits a photon of wavelength λ and returns in the ground state, the principal quantum number of excited state is given by:

0%
$\sqrt{ [(λR-1)/(λR)]}$
0%
$\sqrt{ [(λR)/(λR-1)]}$
0%
$\sqrt{ [λR(λR-1)]}$
0%
$λR(R-1)$

The radius of which of the following orbit is same as that of the first Bohr's orbit of hydrogen atom

0%
${ He }^{ + }(n=2)$
0%
${ Li }^{ 2+ }(n=2)$
0%
${ Li }^{ 2+ }(n=3)$
0%
$\ { Be }^{ 3+ }(n=2)$

λ_{m a x .} = 6563 \dot{A}

$\lambda_{max.} = 6563 \dot{A}$ , then wavelength of second line for Balmer series will be :

0%
$\lambda = \dfrac{16}{3 R}$
0%
$\lambda = \dfrac{36}{5 R}$
0%
$\lambda = \dfrac{4}{3 R}$
0%
none of these

The relation between

λ_{1}

${\lambda}_{1}$ : wavelength of series limit of Lyman series,

λ_{0}

${\lambda}_{0}$ : the wavelength of the series limit of Balmer series and

λ_{3}

${\lambda}_{3}$ : the wavelength of first line of Lyman series is:

0%
${\lambda}_{1}={\lambda}_{2}+{\lambda}_{3}$
0%
${\lambda}_{3}={\lambda}_{1}+{\lambda}_{2}$
0%
${\lambda}_{2}={\lambda}_{3}+{\lambda}_{1}$
0%
None of these

The magnetic field at the center of a hydrogen atom due to the motion of electron varies with the principal quantum number as

0%
$n^{5}$
0%
$n^{3}$
0%
$1/n^{5}$
0%
$1/n^{3}$

According to bohr model, the diameter of first orbit of hydrogen atom will approximately be

0%
$1$ $A^{0}$
0%
$0.529$ $A^{0}$
0%
$2.25$ $A^{0}$
0%
$0.725$ $A^{0}$

Explanation

r = \frac{a_{0} n^{2}}{Z^{2}}

$r=\dfrac{a_{0}n^{2}}{Z^2}$

r = a_{0} (1)^{2}

$r=a_{0}(1)^{2}$

= a_{0}

$=a_{0}$

= 0.53 A^{0}

$=0.53A^{0}$

So,

d = 2 \times r

$d=2\times r$

= 1 A^{0}

$=1\ A^{0}$

How can the brightness of the pattern on the screen or cathode ray tube be changed ?

0%
By changing the negative potential on grid.
0%
By changing the positive potential on grid.
0%
We can't increase the brightness
0%
None of the above

According to kinetic theory of matter, a molecule is the smallest particle of a substance and it possesses :

0%
all the properties of the substance
0%
some of the properties of the substance
0%
none of the properties of the substance
0%
both a and b are true

The time by a photo-electron to come out after the photon strikes is approximately

0%
$10^{-1}s$
0%
$10^{-4}s$
0%
$10^{-10}s$
0%
$10^{-16}s$

Explanation

From experimental aspect, the photoelectric effect is an instantaneous phenomenon. The photon strikes the metallic surface is approximately

10^{- 10}

$10^{-10}$ s.

According to Bohr's theory, the time averaged magnetic field at the centre (i.e nucleus) of a bydrogen atom due to the motion of electrons in the

n^{t h}

$n^{th}$ orbit is propotional to :

(n =

$(n =$ principal quantum number

)

$)$

0%
$n^{-3}$
0%
$n^{-4}$
0%
$n^{-5}$
0%
$n^{-2}$

Explanation

The magnetic field induced at the centre of an orbit due to an electron revolving in the nth orbit of hydrogen atom is given by

B = μ_{0} i_{n / (2 r_{n})}

$B=\mu_0i_{n/(2r_n)}$ ,

Where

i_{n}

$i_n$ is the current in the nth orbit

r_{n}

$r_n$ the nth orbit and

μ_{0}

$\mu_0$ the magnetic permeablity of free space.

r_{n} α n^{2}

$r_n\ \alpha\ n^2$

i_{n} α n^{- 3}

$i_n\ \alpha\ n^{-3}$

Therefore

β α n^{- 5}

$\beta\ \alpha\ n^{-5}$

So the answer is (c)

n^{- 5}

$n^{-5}$ .

The graph which depicts the results of Rutherford gold foil experiment with

α

$\alpha$ -particles is:

θ

$\theta$ : Scattering angle

Y

$Y$ : Number of scattered

α

$\alpha$ -particles detected
(Plots are schematic and not to scale)

Explanation

According to Rutherford's results from the gold foil experiment,

y \propto \frac{1}{\sin^{4} \frac{θ}{4}}

$y \propto \dfrac{1}{\sin^4 \dfrac{\theta}{4}}$

When

\frac{θ}{4} = \frac{π}{2}, θ = 2 π

$\dfrac{\theta}{4} = \dfrac{\pi}{2} , \theta = 2\pi$

From

0

$0$ to

2 π

$2\pi$ ,

\sin^{4} \frac{θ}{4}

$\sin^4 \dfrac{\theta}{4}$ increases.

∴ \frac{1}{\sin^{4} \frac{θ}{4}}

$\therefore \dfrac{1}{\sin^4 \dfrac{\theta}{4}}$ decreases.

Also, at

θ = 0, \sin θ = 0

$\theta = 0, \, \sin \theta = 0$ and

\frac{1}{\sin^{4} \frac{θ}{4}} \to \infty

$\dfrac{1}{\sin^4 \dfrac{\theta}{4}} \to \infty$ .

The matching option is (3)

1477382_1697167_ans_dadf24d1c27a4b06a32ee08601b263cd.png

The transition from the state

n = 4

$n = 4$ to

n = 3

$n = 3$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

0%
$2\rightarrow 1$
0%
$3\rightarrow 2$
0%
$4\rightarrow 2$
0%
$5\rightarrow 4$ .

Explanation

IR corresponds to least value of

(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})

$\displaystyle (\frac{1}{n_1^2}-\frac{1}{n_2^2})$ i.e. from Paschen, Bracket and Pfund series.
Energy of transition is less only in this case.

The acceleration of an electron in the first orbit of the hydrogen atom

(n = 1)

$(n =1)$ is :

0%
$\dfrac{h^2}{\pi^2m^2r^3}$
0%
$\dfrac{h^2}{8\pi^2m^2r^3}$
0%
$\dfrac{h^2}{4\pi m^2r^3}$
0%
$\dfrac {2\pi^2kme^2Z}{\eta h}$

A diatomic molecule has moment of inertia

I

$I$ . By Bohrs quantization condition its rotational energy in the

n^{t h}

$n^{th}$ level (

n = 0

$n = 0$ is not allowed) is

0%
$\displaystyle \frac{1}{\mathrm{n}^{2}}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$
0%
$\displaystyle \frac{1}{\mathrm{n}}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$
0%
$\displaystyle \mathrm{n}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$
0%
$\displaystyle \mathrm{n}^{2}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$

Explanation

L = \frac{n h}{2 π}

$\displaystyle \mathrm{L}=\frac{\mathrm{n}\mathrm{h}}{2\pi}$

K . E . = \frac{L^{2}}{2 I} = (\frac{n h}{2 π})^{2} \frac{1}{2 I}

$\mathrm{K}.\mathrm{E}.\ =\displaystyle \frac{\mathrm{L}^{2}}{2\mathrm{I}}=(\frac{\mathrm{n}\mathrm{h}}{2\pi})^{2}\frac{1}{2\mathrm{I}}$

In a

C O

$\mathrm{C}\mathrm{O}$ molecule, the distance between

C

$\mathrm{C}$ (mass

= 12

$=12$ a.m.u) and

O

$\mathrm{O}$ (mass

= 16

$=16$ a.m.u.), where 1 a.m.u

= \frac{5}{3} \times 10^{- 27} k g,

$=\displaystyle \frac{5}{3}\times 10^{-27} kg,$ is close to

0%
$2.4\times 10^{-10}\mathrm{m}$
0%
$1.9\times 10^{-10}\mathrm{m}$
0%
$1.3\times 10^{-10}\mathrm{m}$
0%
$4.4\times 10^{-11}\mathrm{m}$

Explanation

r_{1} = \frac{m_{2} d}{m_{1} + m_{2}}

$\displaystyle \mathrm{r}_{1}=\frac{\mathrm{m}_{2}\mathrm{d}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$ and

r_{2} = \frac{m_{1} d}{m_{1} + m_{2}}

$\displaystyle \mathrm{r}_{2}=\frac{\mathrm{m}_{1}\mathrm{d}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2}

$\mathrm{I}=\mathrm{m}_{1}\mathrm{r}_{1}^{2}+\mathrm{m}_{2}\mathrm{r}_{2}^{2}$

∴ d = 1.3 \times 10^{- 10} m

$\therefore \mathrm{d}=1.3\times 10^{-10}\mathrm{m}$ .

The electric potential between a proton and an electron is given by

V = V_{0} \ln \frac{r}{r_{0}}

$V=\displaystyle \mathrm{V}_{0}\ln\frac{\mathrm{r}}{\mathrm{r}_{0}}$ , where

r_{0}

$\mathrm{r}_{0}$ is a constant. Assuming Bohr's model to be applicable, write variation of

r_{n}

$\mathrm{r}_{\mathrm{n}}$ with

n, n

$\mathrm{n},\ \mathrm{n}$ being the principal quantum number?

0%
$\mathrm{r}_{\mathrm{n}}\propto \mathrm{n}$
0%
$\mathrm{r}_{\mathrm{n}}\propto 1/\mathrm{n}$
0%
$\mathrm{r}_{\mathrm{n}}\propto \mathrm{n}^{2}$
0%
$\mathrm{r}_{\mathrm{n}}\propto 1/\mathrm{n}^{2}$

Explanation

Electric potential between proton and electron in nth orbit is given as:

V = V_{o} l n \frac{r_{n}}{r_{o}}

$V = V_o ln\dfrac{r_n}{r_o}$

Thus the coulomb force

| F_{c} | = e \frac{d V}{d r} = e \frac{V_{o}}{r_{n}}

$|F_c| = e\dfrac{dV}{dr} = e\dfrac{V_o}{r_n}$

This coulombic force is balance by the centripetal force.

∴ \frac{m v^{2}}{r_{n}} = e \frac{V_{o}}{r_{n}} ⟹ v = \sqrt{\frac{e V_{o}}{m}}

$\therefore \dfrac{mv^2}{r_n} = e\dfrac{V_o}{r_n} \implies v = \sqrt{\dfrac{e V_o}{m}}$ means that speed is constant

Now from

m v r_{n} = \frac{n h}{2 π} ⟹ r_{n} \propto n

$m v r_n = \dfrac{nh}{2\pi} \implies r_n \propto n$ (as

v = c o n s t a n t

$v = constant$ )

Match the appropriate pairs from Lists I and II:

List-I	List - II
a) Nitrogen molecule	e) Continuous spectrum
b) Incandescent solids	f) Absorption spectrum
c) Fraunhofer lines	g) Band spectrum
d) Electric arc between iron rods	h) Emission spectrum

0%
a-g, b-e, c-f, d-h
0%
a-f, b-e, c-h, d-g
0%
a-h, b-e, c-f, d-g
0%
a-e, b-g, c-h, d-f

The electrostatic energy of

Z

$Z$ protons uniformly distributed throughout a spherical nucleus of radius

R

$R$ is given by

E = \frac{3}{5} \frac{Z (Z - 1) e^{2}}{4 π ε_{0} R}

$\displaystyle E=\frac { 3 }{ 5 } \frac { Z\left( Z-1 \right) { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }R }$
The measured masses of the neutron,

_{1}^{1} H,_{7}^{15} N

$\displaystyle _{ 1 }^{ 1 }{ H },\, _{ 7 }^{ 15 }{ N }$ and

_{8}^{15} O

$\displaystyle _{ 8 }^{ 15 }{ O }$ are

1.008665 u, 1.007825 u, 15.000109 u

$1.008665\ u, 1.007825 \ u, 15.000109\ u$ and

15.003065 u

$15.003065\ u$ , respectively. Given that the radii of both the

_{7}^{15} N

$\displaystyle _{ 7 }^{ 15 }{ N }$ and

_{8}^{15} O

$\displaystyle _{ 8 }^{ 15 }{ O }$ nuclei are same,

1 u = 931.5 M e V / c^{2}

$\displaystyle 1u=931.5\quad { MeV }/{ { c }^{ 2 } }$ (c is the speed of light) and

e^{2} / (4 π ε_{0}) = 1.44 M e V f m

$\displaystyle { { e }^{ 2 } }/{ \left( 4\pi { \varepsilon }_{ 0 } \right) }=1.44\,MeV\,fm$ . Assuming that the difference between the binding energies of

_{7}^{15} N

$\displaystyle _{ 7 }^{ 15 }{ N }$ and

_{8}^{15} O

$\displaystyle _{ 8 }^{ 15 }{ O }$ is purely due to the electrostatic energy, the radius of either of the nuclei is

(1 f m = 10^{- 15} m)

$(1\, fm = 10^{-15}m)$ :

0%
$2.85\ fm$
0%
$3.03\ fm$
0%
$3.42\ fm$
0%
$3.80\ fm$

Explanation

For

_{7}^{15} N

$^{15}_{7} N$ :

Z_{1} = 7

$Z_1 = 7$

N_{1} = 8

$N_1 = 8$

For

_{8}^{15} O

$^{15}_{8} O$ :

Z_{2} = 8

$Z_2 = 8$

N_{2} = 7

$N_2 = 7$

Difference in the electrostatic energy,

Δ E = E_{2} - E_{1}

$\Delta E = E_2 -E_1$

Δ E = \frac{3}{5} (\frac{8 (8 - 1)}{R} - \frac{7 (7 - 1)}{R}) \times 1.44 = \frac{12.096}{R}

$\Delta E = \dfrac{3}{5} \bigg(\dfrac{8 (8-1)}{R} - \dfrac{7(7-1)}{R}\bigg) \times 1.44 = \dfrac{12.096}{R}$

M e V

$MeV$

f m

$fm$

Mass defect,

Δ M = Z M_{H} + N M_{n} - M_{a t o m}

$\Delta M = Z M_H + N M_n - M_{atom}$

∴

$\therefore$ For

_{7}^{15} N

$^{15}_{7} N$

Δ M_{1} = 7 (1.007825) + 8 (1.008665) - 15.000109 = 0.12399

$\Delta M_1 = 7(1.007825) + 8(1.008665) - 15.000109 = 0.12399$ u

For

_{8}^{15} O

$^{15}_{8} O$ ,

Δ M_{2} = 8 (1.007825) + 7 (1.008665) - 15.003065 = 0.12019

$\Delta M_2 = 8(1.007825) + 7(1.008665) - 15.003065 = 0.12019$ u

Difference in binding energy,

Δ B . E = [Δ M_{1} - Δ M_{2}] \times 931.5

$\Delta B.E = [\Delta M_1 - \Delta M_2] \times 931.5$

M e V

$MeV$

⟹

$\implies$

Δ B . E = [0.12399 - 0.12019] \times 931.5 = 3.54

$\Delta B.E = [0.12399 -0.12019] \times 931.5 = 3.54$

M e V

$MeV$

According to the question,

\frac{12.096}{R} = 3.54

$\dfrac{12.096}{R} = 3.54$

⟹ R = 3.42

$\implies R = 3.42$

f m

$fm$

For which one of the following, Bohr model is not valid?

0%
Singly ionised helium atom $(He^+)$
0%
Deuteron atom
0%
Singly ionised neon atom $(Ne^+)$
0%
Hydrogen atom

The value of Planck's constant is

6.63 \times 10^{- 34} J s

$6.63\times 10^{-34}Js$ . The speed of light

3 \times 10^{17} n m s^{- 1}

$3\times 10^{17}\:nm\:s^{-1}$ . Which value is closest to the wavelength in nanometer of a quantum of light with frequency of

6 \times 10^{15} s^{- 1}

$6\times 10^{15}s^ {-1}$ ?

0%
$10$
0%
$25$
0%
$50$
0%
$75$

Explanation

v = \frac{c}{λ}

$\displaystyle v=\frac{c}{\lambda }$

∴ λ = \frac{3 \times 10^{17} n m s^{- 1}}{6 \times 10^{15} s^{- 1}} = 50 n m

$\therefore \displaystyle \lambda =\frac{3\times 10^{17}nms^{-1}}{6\times 10^{15}s^{-1}}=50\:nm$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
0%
Assertion is correct but Reason is incorrect.
0%
Both Assertion and Reason are incorrect.

Whenever a stream of electrons collides with a stream of photons, in this collision, which of the following is not conserved?

0%
Linear momentum
0%
Total energy
0%
No. of photons
0%
No. of eletrons

What would be the radius of second orbit of

{H e}^{+}

${He}^{+}$ ion?

0%
$1.058\mathring { A }$
0%
$3.023\mathring { A }$
0%
$2.068\mathring { A }$
0%
$4.458\mathring { A }$

Explanation

Radius of the nth orbit,

r_{n} = 0.529 \frac{n^{2}}{Z}

$r_n = 0.529 \dfrac{n^2}{Z}$

A^{o}

$A^o$

For

H e^{+}

$He^+$ ion,

Z = 2

$Z = 2$

Radius of 2nd orbit,

r_{2} = 0.529 \frac{2^{2}}{2}

$r_2 = 0.529\dfrac{2^2}{2}$

⟹ r_{2} = 1.058

$\implies r_2 = 1.058$

\overset{˚}{A}

$\mathring A$

According to Bohr's theory, the radius of the $n^{th}$ orbit of an atom of atomic number $Z$ is proportional to

0%
$\dfrac{n^2}{Z^2}$
0%
$\dfrac{n^2}{Z}$
0%
$\dfrac{n}{Z}$
0%
$n^2Z^2$

Explanation

According to Bohr's theory, the coulomb force on the electron causes its centripetal acceleration.

⟹ \frac{1}{4 π ϵ_{0}} \frac{Z e^{2}}{r^{2}} = \frac{m v^{2}}{r}

$\implies \dfrac{1}{4\pi\epsilon_0}\dfrac{Ze^2}{r^2}=\dfrac{mv^2}{r}$

Also one of Bohr's postulates states

m v r = \frac{n h}{2 π}

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

Thus eliminating

v

$v$ from the equations gives

⟹ r \propto \frac{n^{2}}{Z}

$\implies r\propto \dfrac{n^2}{Z}$

Consider the following two statements A and B and identify the correct choice in the given answers
A) Line spectra is due to atoms in gaseous state
B) Band spectra is due to molecules

0%
Both A and B are false
0%
A is true but B is false
0%
A is false but B is true
0%
Both A and B are true.

Match the following :

List - 1	List - 2
a) Burning candle	e) line spectrum
b) Sodium vapour	f) continuous spectrum
c) Bunsen flame	g) bond spectrum
g) Dark lines in solar spectrum	h) Absorption spectrum

0%
a-g, b-e, c-f, d- h
0%
a- g, b-f, c-e, d-h
0%
a-f, b-g, c-e, d-h
0%
a-f, b-e, c-g, d-h

An incandescent filament emits a spectrum which is :

0%
line spectrum
0%
band spectrum
0%
continuous spectrum
0%
characteristic spectrum

Light from a tungsten filament lamp gives

0%
Absorption spectrum
0%
Emission spectrum
0%
Atomic spectrum
0%
Discontinuous spectrum

Check the wrong statement :

0%
Line spectrum is characteristic of the element
0%
Absorption line spectrum is characteristic of the element
0%
Continuous spectrum is characteristic of the source of light
0%
There are two prominent yellow lines in the spectrum of sodium

Solar spectrum is an example of

0%
line emission spectrum
0%
band absorption spectrum
0%
line absorption spectrum
0%
continuous emission spectrum

Which of the following scientists developed the nuclear model of the atom ?

0%
John Dalton
0%
Robert Milikan
0%
Henry Moseley
0%
Ernest Rutherford

The spectra used to identify the elements in the mixture is :

0%
Emission
0%
Absorption
0%
Emission and Absorption
0%
Molecular spectrum

The element which was observed in solar spectrum is

0%
Helium
0%
Xenon
0%
Neon
0%
Argon

The rest mass of a photon is

0%
zero
0%
$1.6\times 10^{-19}kg$
0%
$3.1\times 10^{-30}kg$
0%
$9.1\times 10^{-31}kg$

Bohrs atomic model assumes :

0%
the nucleus is of infinite mass and is at rest
0%
electron in a quantized orbit will not radiate energy
0%
mass of the electron remains constant
0%
all of the above

The incorrect statement from the following is:

0%
Material wave (de-Broglie wave) can travel in vacuum.
0%
Electromagnetic wave can travel through vacuum.
0%
The velocity of photon is the same as light passes through any medium.
0%
Wavelength of de-Broglie wave depends upon velocity.

Explanation

The speed at which light propagates through medium depends on the refractive index of the material (n). It is given by:

v = c / n

$v=c/n$ .

The particle that possesses half integral spin is

0%
Photon
0%
Pion
0%
Proton
0%
K-meson

CBSE Questions for Class 12 Medical Physics Atoms Quiz 1 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 1 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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