MCQ Questions for Class 12 Medical Physics Atoms Quiz 11

The ratio of the frequencies of the long wavelength limits of the Lyman and Balmer series of hydrogen is:

0%
$27 : 5$
0%
$5 : 27$
0%
$4 : 1$
0%
$1 : 4$

The anode voltage of a photocell is kept fixed. The wavelength (i) of light falling on the cathode is gradually changed. The plate current (I) of the photocell varies as

Explanation

$\lambda$ is increased, there will be a value of

$\lambda$ above which photo electrons will cease to come out. So, photo current will be zero.

The shortest wavelength in the Lyman series of hydrogen spectrum is 912 A corresponding to a photon energy of 13.6 eV. The shortest wavelength in the Balmer series is:

0%
$3648 A^o$
0%
$228 A^o$
0%
$6566 A^o$
0%
$8208 A^o$

Rutherford's experiment on scattering of particles showed for the first time that the atom has:

0%
electrons
0%
protons
0%
nucleus
0%
neutrons

From lyman series of Pfund series, frequency__________

0%
increase
0%
decrease
0%
increase or decrease
0%
no change

Explanation

In physics and chemistry the lyman series is a hydrogen spectral series transitions and resulting ultraviolet emission lines of the hydrogen atom as electron goes from

$n \geqslant 2\,to\,n = 1.$

$,$ lyman series of pfund series

$,$ frequency is decrease

$.$

hence,

option

$(B)$ is correct answer.

In the Geiger - Marsden scattering experiment, find the distance of closest approach to the nucleus of a

$7.7Mev$

$\alpha$ - particle before it comes momentaily to rest and reverses its direction. (

$'Z'$ for gold nucleus

$= 79$ )

0%
$10fm$
0%
$20fm$
0%
$30fm$
0%
$40fm$

In hydrogen atom, an electron makes transition from

$n=3$ to

$n=2$ state in time interval of

$1.2\times{10}^{-8}s$ . Calculate the average torque (in Nm) acting on the electron during this transition.

0%
$1.055\times{10}^{-26}$
0%
$4.40\times{10}^{-27}$
0%
$1.7\times{10}^{-26}$
0%
$8.79\times{10}^{-27}$

Explanation

Given,

$t=1.2\times 10^{-8}sec$

By Bohr's model,

Angular momentum,

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

$n=3$ ,

$L_3=\dfrac{3h}{2\pi}$

$n=2$ ,

$L_2=\dfrac{2h}{2\pi}$

Torque is the rate of change of angular momentum

$\tau=\dfrac{L_2-L_3}{t}$

$\tau=\dfrac{(2h/2\pi)-(3h/2\pi)}{t}$

$|\tau|=\dfrac{h}{2\pi t}$

$|\tau|=\dfrac{6.65\times 10^{-34}}{2\times 3.14\times 1.2\times 10^{-8}}$

$|\tau|=1.055\times 10^{-26}Nm$

The correct option is A.

Time taken for an electron to complete one revolution in the Bohr orbit of hydrogen atom is:

0%
$\dfrac { 4\pi ^{ 2 }mr^{ 2 } }{ nh }$
0%
$\dfrac { nh }{ 4{ \pi }^{ 2 }mr }$
0%
$\dfrac { nh }{ 4{ \pi }^{ 2 }mr^{ 2 } }$
0%
$\dfrac { h }{ 2\pi mr }$

Explanation

By Bohr postulate,

$mvr = n \dfrac{h}{2 \pi}$ or

$v = \dfrac{nh}{2 \pi m r }$

No. of revolutions per sec

$= \dfrac{{\text{Velocity}}}{{\text{Circumference of the orbit}}}$

$= \dfrac{v}{2 \pi r } = \dfrac{nh}{2 \pi m r} \times \dfrac{1}{2 \pi r} = \dfrac{nh}{4 \pi^2 mr^2}$

Time taken for one revolution

$= \dfrac{4 \pi^2 mr^2}{nh}$

In which of the following system will the radius of the

$2^{nd}$ orbit be minimum?

0%
$H$
0%
$Mg^{+2}$
0%
$He^+$
0%
$B$

According to Bohr hydrogen atom model the radius of stationary orbit directly proportional to principal quantum number n as

$\rightarrow$

0%
$\dfrac { 1 }{2 n }$
0%
$n^2$
0%
${ n }$
0%
none of these

The energy of electron in the excited state of hydrogen atom is

$-0.85 eV$ . If 'h' is Plank's constant, the angular momentum of electron in the excited state is :

0%
$\dfrac{4h}{2 \pi}$
0%
$h$
0%
$\dfrac{3 h}{2 \pi}$
0%
$\dfrac{h}{\pi}$

Explanation

Energy of the

$nth$ state of H-atom

$E_n = \dfrac{-13.6}{n^2} \ eV$

So,

$-0.85 = -\dfrac{13.6}{n^2}$

$n^2 = \dfrac{13.6}{0.85} = 16$

$\implies \ n = 4$

Angular momentum

$L = \dfrac{nh}{2\pi} = \dfrac{4h}{2\pi}$

The nuclear radius of a certain nucleus is

$7.2 f m$ and it has a charge of

$1.28 \times 10 ^ { - 17 } { C }.$ The number of neutrons inside the nucleus is

0%
$136$
0%
$140$
0%
$126$
0%
$142$

Explanation

$\begin{array}{l} { n_{ p } }+{ n_{ p } } \\ A={ \left( { \frac { r }{ { { r_{ o } } } } } \right) ^{ 3 } } \\ ={ \left( { \frac { { 7.2 } }{ { 1.2 } } } \right) ^{ 3 } }={ 6^{ 3 } }=216 \\ { x_{ p } }=\frac { q }{ e } =\frac { { 1.28\times { { 10 }^{ -17 } } } }{ { 1.6\times { { 10 }^{ -19 } } } } =80 \\ \therefore \, \, { n_{ p } }=216-80=136 \end{array}$

Two monochromatic light sources,

$A$ and

$B$ , emit the same number of photons per second. The wavelength of

$A$ is

${\lambda}_{A}=400nm$ and that

$B$ is

${\lambda}_{B}=600nm$ . The power radiated by source

$B$ is

0%
equal to that of source $A$
0%
less than that of source $A$
0%
greater than that of source $A$
0%
cannot be compared to that from source $A$ using the available data.

The ratio of the kinetic energy to the total energy of an electron in a Bohr orbit is

0%
$1:-1$
0%
$2$
0%
$1 : 2$
0%
$None$

Explanation

$k =kinetic\, energy$

$E=total\, energy$

$K=-E$

$\dfrac{K}{E}=\dfrac{1}{-1}$

A certain dye absorbs light of

$\lambda =4000\overset { \circ }{ A }$ and emits fluorescent light of

$\lambda =5000\overset { \circ }{ A }$ . Assuming that under given condition 50% of the absorbed energy is re-emitted out fluorescence, Calculate the ratio of the number of quanta emitted out to the number of quanta absorbed :

0%
$\dfrac { 5 }{ 8 }$
0%
$\dfrac { 8 }{ 5 }$
0%
$\dfrac { 3 }{ 8 }$
0%
$\dfrac { 8 }{ 3 }$

According to Bohr's hypothesis , the angular momentum of the electron on any stationary orbit of radius r is proportional to ......

0%
independent of r
0%
1/r
0%
$\sqrt r$
0%
$r^2$

Explanation

We know, there will be two forces acting, electrostatic and centripetal force; so

$F_{c}=F_{e} \\$

$\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} x \frac{e^{2}}{r^{2}} \\$

$v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{m r} \\$

$V=\frac{e^{2}}{4 \pi \varepsilon_{0} r}$

We know,

$r_{n} \propto \frac{n^{2}}{2}$

$n=(z r)^{1 / 2} \\$

$L=\frac{n h}{2 \pi} \times n \times(22)^{1 / 2}$

$\therefore \quad L \alpha \sqrt{r}$

(c)

$\sqrt{2}$

Find the average of radius for $A{u^{197}}$ .

0%
$6.4\;{\text{fm}}$
0%
$4.6\;{\text{fm}}$
0%
$0.46\;{\text{fm}}$
0%
None of these

The size of an atom is of the order of

0%
$10^{-8} m$
0%
$10^{-10} m$
0%
$10^{-12} m$
0%
$10^{-14} m$

A photon and an electron both have wavelength 1A The ratio of energy of photon to that of electron is

0%
1
0%
0.012
0%
82.7
0%
$10^{-10}$

Wavelength of the first line of Balmer series is

$600nm$ . The wavelength of second line of the Balmer series will be:

0%
$444nm$
0%
$800nm$
0%
$388nm$
0%
$632nm$

Explanation

$\dfrac{1}{\lambda_1} = R \left[\dfrac{1}{2^2} - \dfrac{1}{3^2}\right] = \dfrac{5}{36}R$

$\dfrac{1}{\lambda_2} = R\left[\dfrac{1}{2^2} - \dfrac{1}{4^2} \right] = R \left[\dfrac{3}{16}\right]$

$\dfrac{\lambda_2}{\lambda_1} = \dfrac{5R/36}{3R/16} = \dfrac{5}{36}\times \dfrac{16}{3}$

$\lambda_2 = \dfrac{5}{36}\times \dfrac{16}{3}\times 660 = 444.44nm$

$H$ -atm spectrum

$V$ is the wave number

$V_1 = V_{min} + V_{max}$ for Lyman series

$V_1 = V_{min} + V_{max}$ for Balmer series then

$V_1 : V_2$

0%
$9:2$
0%
$3:2$
0%
$5:2$
0%
$7:2$

Explanation

$V_1 = R (1)^2 \left[\dfrac{1}{1^2} - \dfrac{1}{2^2}\right] + R(1)^2 \left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right) = R Z^2\left(\dfrac{7}{4}\right)$

$V_2 = R (1)^2 \left[\dfrac{1}{2^2} - \dfrac{1}{3^2}\right] + R(1)^2 \left(\dfrac{1}{2^2} - \dfrac{1}{\infty^2}\right) = R (1)^2\left(\dfrac{7}{18}\right)$

$V_1 : V_2 = 9 : 2$

An electron of H-atom de-excites from energy level

$n_1=2$ to

$n_2=1$ and the emitted photon is incident on

$He^+$ ions in ground and first excited state. Which of following transition is possible.

0%
$n=1$ to $n=4$
0%
$n=2$ to $n=4$
0%
$n=2$ to $n=3$
0%
$n=1$ to $n=3$

Explanation

$E=13.6\times 1(\dfrac { 1 }{ 1^{ 2 } } -\dfrac { 1 }{ 2^ 2 })$

$\Rightarrow10.2eV$

for He+in

$n=2$

$10.2eV=13.6\times4(\dfrac { 1 }{ 2^{ 2 } } -\dfrac { 1 }{ n^ 2 } )$

$\Rightarrow n=4$

The energy level diagram for an hydrogen like atom is shown in the figure.The radius of its first Bohr orbit is
0 eV ____________n =

$\infty$
-6.04 eV __________ n = 3
-13.6 eV ____________ n =2
-54.4 eV ____________ n=1

0%
$0.265 \mathring { A }$
0%
$0.53 \mathring { A }$
0%
$0.132 \mathring { A }$
0%
None of these

Explanation

We know that

$E_n=-13.6 \dfrac{Z^2}{n^2}eV$ and

$r_n=0.53 \dfrac{n^2}{Z}(A^o)$

Here for

$n=1,\ E=-54.4\ eV$

Therefore

$-54.4=-13.6 \dfrac{Z^2}{1^1} \Rightarrow Z=2$

hence radius of first orbit

$r=\dfrac{0.53(1)^2}{2}=0.265\ A^o$

$'\lambda '$ the longest wavelength of Lyman series, then shortest wavelength line that of the Balmer series is:-

0%
$5 \lambda$
0%
$2 \lambda$
0%
$4 \lambda$
0%
$3 \lambda$

The number density of molecules of a gas depends on their distance

$r$ from the origin as,

$n(r) = n_0 e^{-\alpha r^4}$ . Then the total number of molecules is proportional to:

0%
$n_0 \alpha^{1/4}$
0%
$n_0 \alpha^{-3}$
0%
$n_0 \alpha^{-3/4}$
0%
$\sqrt{n_0}\alpha^{1/2}$

Explanation

Given number density of molecules of gas as a function of

$r$ is

$n(r) = n_0 e^{-\alpha r^4}$

$\therefore$ Total number of molecule

$= \overset{\infty}{\underset{0}{\int}} n(r) dV$

$= \overset{\infty}{\underset{0}{\int}} n_9 e^{-\alpha r^4} 4\pi r^2 dr$

$\therefore$ Number of molecules is proportional to

$n_0 \alpha^{-3/4}$

According to Bohr's theory, the product of velocity of electron in

$n^{th}$ orbit and the radius of

$n^{th}$ orbit is:

0%
inversely proportional to the mass of the electron
0%
directly proportional to the mass of the electron
0%
independent of the mass of the electron
0%
directly proportional to the square of mass of the electron

Taking the wavelength of first Balmer line in hydrogen spectrum (

$n=3$ to

$n=2$ ) as

$660nm$ , the wavelength of the 2nd Balmer line (

$n=4$ to

$n=2$ ) will be:

0%
$889.2nm$
0%
$642.7nm$
0%
$488.9nm$
0%
$388.9nm$

Explanation

$\cfrac { 1 }{ 660 } =R\left( \cfrac { 1 }{ { 2 }^{ 2 } } -\cfrac { 1 }{ { 3 }^{ 2 } } \right) =\cfrac { 5R }{ 36 } ....(1)\quad$

$\cfrac { 1 }{ \lambda } =R\left( \cfrac { 1 }{ { 2 }^{ 2 } } -\cfrac { 1 }{ { 4 }^{ 2 } } \right) =\cfrac { 3R }{ 16 } ....(2)\quad \quad$
dividing (1) with (2)

$\cfrac { \lambda }{ 660 } =\cfrac { 5\times 16 }{ 36\times 3 }$

$\lambda =\cfrac { 4400 }{ 9 } =488.88=488.9nm$

The ratio of mass densities of nuclei of

$^{40} Ca$ and

$^{16} O$ is close to :

0%
$1$
0%
$2$
0%
$0.1$
0%
$5$

Explanation

$\rho= \dfrac{Mass}{Volume}$

$r = r_0 A^{\dfrac {1}{3}}$

r = radius of nucleus

A = Atomic number

$r_0 = constant$

$\rho = \dfrac{Mass}{\dfrac{4}{3}\pi r^3}$

$\rho \propto \dfrac{Mass(Z)}{A}$

For, Oxygen, Z=16, A=8

Calsium, Z=40, A=20

$\dfrac{\rho_{o_{2}}}{\rho_{ca}}=\dfrac{Z_{o_{2}}}{Z_ca}=\dfrac{16}{8}\times \dfrac{20}{40} = \dfrac{1}{1}$

$Li^{++}$ , electron in first Bohr orbit is excited to a level by a radiation of wavelength

$\lambda$ . when the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of

$\lambda$ ?
(Given :

$h = 6.63 \times 10^{34} Js$ ;

$\, c = 3 \times 10^8 ms^{-1}$ )

0%
$9.4 nm$
0%
$12.3 nm$
0%
$10.8 nm$
0%
$11.4 nm$

Explanation

$\Delta E = \dfrac{hc}{\lambda}$

$13.6 \times 9 - 0.85 \times 9 = \dfrac{hc}{\lambda}$

$\lambda = \dfrac{hc}{9 \times (13.6 - 0.85) eV}$

$= \dfrac{1240 eV . nm}{9 \times 12.75 eV}$

$\lambda = 10.8 nm$
1247495_1614656_ans_8c4b1a0fff78400eb0b790ee037407c5.PNG

In a muonic atom, a muon of mass of 200 times of that of electron and same charge is bond to the proton. The wavelengths of its Balmer series are in the range of :

0%
X-rays
0%
infrared
0%
$\gamma$ rays
0%
microwaves

Explanation

Energy of an orbit,

$E=\dfrac{-me^4}{8\in_0^2h^2}\dfrac{z^2}{n^2}$

$\because E$ increase 200 times

$\therefore \lambda$ will decrease by 200 times.

Hence

$\lambda$ will be in the range of x-rays.

The total energy of an electron revolving in the second orbit of hydrogen atom is?

0%
$-13.6$ eV
0%
$-1.51$ eV
0%
$-3.4$ eV
0%
Zero

Explanation

$E_n=\dfrac{-13.6z^2}{n^2}$

$Z=1; n=2; E_n=-3.4eV$ .

Which of the following is not a correct statement according to Rutherford's atomic model?

0%
$99\%$ of mass of an atom is centered in the nucleus.
0%
Most of the part inside the atom is empty.
0%
The size of nucleus is very small in comparison to the atom.
0%
Electrons revolve around the nucleus in a fixed circular path.

An experiment in which

$\alpha$ - particles were deflected by a gold foil produced new insights into the structure of the atom

Which conclusion can be drawn from the results of the experiment?

0%
Atomic nuclei occupy a very small fraction of the volume of an atom
0%
Electrons orbit the atomic nucleus
0%
Some atoms of the same element contain different numbers of neutrons
0%
The atomic nucleus contains protons and neutrons

Explanation

This experiment is said to be Rutherford experiment.

In this experiment,most of the

$\alpha$ particle get passes through the gold foil . Only a small fraction of

$\alpha$ particle get deflected by it.

This implies that the most of the mass of the atom is concentrated st the nucleus i.e. at the centre of atom.

So, it is concluded that a nuclei occupy a very small fraction of the volume of an atom

A student measures the level of radiation emitted by a radioactive sample.
The table shows the readings she records on the counter over a short period of time.

Counter reading/ counts per minute

$106$

$96$

$98$

$100$

The sample is removed and the counter then shows a background radiation reading of

$4$ counts per minute.
What is the best estimate for the average count rate due to the radioactive sample?

0%
$96$ counts per minute
0%
$98$ counts per minute
0%
$100$ counts per minute
0%
$104$ counts per minute

Explanation

Total count per minute =

$106 +96 + 98 +1 100 = 400$ counts

so, Average count of sample

$\dfrac {400}{4} = 100$

net count per minute = Average count - number of sample

$100 - 4 = 96$ counts per minute

The period of revolution of an electron in the ground state of hydrogen atom is T. The period of revolution of the electron in the first excited state is?

0%
$2$ T
0%
$4$ T
0%
$6$ T
0%
$8$ T

Explanation

$T^2\alpha R^3\alpha n^6\Rightarrow T\alpha n^3$
Hence

$\dfrac{T^1}{T}=2^3=8$ .

An excited

$He^+$ ion emits two photons in succession, with wavelengths

$108.5 nm$ and

$30.4 nm$ , in making a transition to ground state. The quantum number

$n$ , corresponding to its initial excited state is (for photon of wavelength

$\lambda$ , energy

$E = \dfrac{1240 eV}{\lambda (in\, nm)}$ )

0%
$n = 5$
0%
$n = 4$
0%
$n = 6$
0%
$n = 7$

Explanation

$\dfrac{1}{\lambda} = R \left(\dfrac{1}{m^2} - \dfrac{1}{n^2}\right)z^2$

$\dfrac{1}{1085} = R \left(\dfrac{1}{m^2} - \dfrac{1}{n^2}\right)^2$

$\dfrac{1}{304} = R \left(\dfrac{1}{1^2} - \dfrac{1}{m^2}\right)2^2$

$\therefore m = 2$

$\therefore n = 5$

The diagram shows a cathode-ray oscilloscope display of an electromagnetic wave. The time base setting is

$0.20 \mu s cm^{-1}$ . Which statement is correct?

0%
The frequency of the wave is 2.5 MHz and it lies in the microwave region of the
electromagnetic spectrum
0%
The frequency of the wave is 2.5 MHz and it lies in the radio-wave region of the
electromagnetic spectrum.
0%
The frequency of the wave is 5.0 MHz and it lies in the microwave region of the
electromagnetic spectrum.
0%
The frequency of the wave is 5.0 MHz and it lies in the radio-wave region of the
electromagnetic spectrum.

Explanation

It is given that

$\dfrac{\lambda}{2}= 1.0\ cm$

$\implies \lambda = 2.0\ cm$

time base =

$0.20\mu\ scm^{-1}$

So,

Time period =

$0.20\mu\ cms^{-1}\times \lambda$

$\implies T=0.4\mu\ s$

So, frequency =

$\dfrac{1}{T}$

$\implies f = \dfrac{1}{0.40\times 10^{-6}}\ s^{-1}$

$\implies f = 2.5MHz$

Since the frequency range of radio range is

So, the given wave is in radio-wave region of spectrum.

The number of nodal sphere of

$3s-$ orbital are

0%
$1$
0%
$2$
0%
$3$
0%
$0$

Cathode rays are made to pass between the plates of a charged capacitor, It attracts.

0%
Towards positive plate
0%
Towards negative plate
0%
(a) and (b) are correct
0%
(a) and (b) are wrong

A moving electron has numerical relation

$\lambda = h$ . Then

0%
$m_{e} = \dfrac {1}{v_{e}}$
0%
$v_{e} = \dfrac {1}{m_{e}}$
0%
Both (a) and (b)
0%
None of these

The cathode ray particles originates in a discharge tube from the

0%
Cathode
0%
Anode
0%
Source of high voltage
0%
Residual gas

In Rutherford experiment the number of

$\alpha -$ particles scattered through an angle

$60^o$ is

$112$ per minute, then the number of

$\alpha -$ particles scattered through an angle of

$90^o$ per minute by the same nucleus is :

0%
$28$ per min
0%
$112$ per min
0%
$12.5$ per min
0%
$7$ per min

Which of the following is incorrect regarding Rutherford's atomic model ?

0%
Atom contains nucleus
0%
Size of nucleus is very small in comparison to that of atom
0%
Nucleus contains about $90\%$ mass of the atoms
0%
Electrons revolve round the nucleus with uniform speed

For which of the following set of quantum numbers, an electron will have the lowest energy ?

0%
$3,2,1,\dfrac {1}{2}$
0%
$4,2,-1,\dfrac {1}{2}$
0%
$4,1,0,-\dfrac {1}{2}$
0%
$5,0,0,\dfrac {1}{2}$

Which is the common node for all orbitals:

0%
$x, y$ and $z-$ axis
0%
$xy, yz$ and $xz$ planes
0%
nodal spheres between two orbitals
0%
nucleus

The angular momentum of the

$\alpha-$ particles which are scattered through large by the heavier nuclei, is conserved because of the

0%
nature of repulsive force
0%
conservation of kinetic energy
0%
conservation of potential energy
0%
There is no external torque.

Which of the following curves may represent radius of orbit in

$H-$ atom as a function quantum number ?

The number of orbitals in

$3^{rd}$ orbit are :

0%
$3$
0%
$10$
0%
$18$
0%
$none\ of\ these$

The order of size of nucleus and Bohr radius of an atom respectively are :

0%
$10^{-14}m, 10^{-10}m$
0%
$10^{-10}m, 10^{-8}m$
0%
$10^{-20}m, 10^{-16}m$
0%
$10^{-8}m, 10^{-6}m$

The radius of hydrogen atom, when it is in its second excited state, becomes :

0%
half
0%
double
0%
four times
0%
nine times

Explanation

The relation between the radius and the orbit of the atom is:

$r\propto n^2$

For second excited state,

$n=3$ .

$r \propto (3)^2$

$\Rightarrow r\propto 9$

CBSE Questions for Class 12 Medical Physics Atoms Quiz 11 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 11 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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