Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Light of wavelength $\lambda$ strikes a photo sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$, then the wavelength must be changed to $\lambda'$, where :

  1. $\lambda' > \lambda$

  2. $\lambda' = \dfrac {\lambda}{2}$

  3. $\lambda' = 2\lambda$

  4. $\dfrac {\lambda}{2} > \lambda' > \lambda$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have, $E _{k} = \dfrac {hc}{\lambda} - \phi _{0}$


and $2E _{k} = \dfrac {hc}{\lambda'} - \phi _{0}$

By the two relations, we have

$\lambda' > \dfrac {\lambda}{2}$

and $\lambda' < \lambda$

So, $\dfrac {\lambda}{2} < \lambda' < \lambda$.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A photo-cell is illuminated by a source of light, which is placed at a distance $d$ from the cell. If the distance become $d/2$, then number of electrons emitted per second will be : 

  1. Remain same

  2. Four times

  3. Two times

  4. One-fourth
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Intensity, $I=\dfrac { E }{ At } \ $


where E is the total energy of source,


A is area of illuminated surface,

t is time

$\therefore I=\dfrac { E }{ 4\pi { { { r }^{ 2 } }t } } \\ \dfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\dfrac { E }{ 4\pi { { { r } _{ 1 }^{ 2 } }t } } \times \dfrac { 4\pi { { { r } _{ 2 }^{ 2 } }t } }{ E } \\ =\dfrac { { r } _{ 2 }^{ 2 } }{ { r } _{ 1 }^{ 2 } } \\ \dfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\dfrac { 4 }{ 1 } \\ \Rightarrow \dfrac { { I } _{ 2 } }{ { I } _{ 1 } } =\dfrac { 1 }{ 4 } $

Since, $\propto$ number of photoelectrons emitted

Therefore, Number of electrons emitted is a quarter of the initial number.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The formula for kinetic mass of photon is:
where $h$ is Planck's constant and $v,\lambda, c$ are frequency, wavelength and speed of photon respectively.

  1. $\cfrac { hv }{ \lambda } $

  2. $\cfrac { h }{ c\lambda } $

  3. $\cfrac { hv }{ c } $

  4. $\cfrac { h\lambda }{ c } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy of the photon is given by,

$E=hv=\dfrac{hc}{\lambda}$,  ...(1)
also for particle of zero mass,
$E=mc^2$,   ...(2)
form eq(1) and (2) we get,
$mc^2=\dfrac{hc}{\lambda}$
$\therefore$ kinetic mass of photon$= m=\dfrac{h}{c\lambda}$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

If momentum of a photon of an electromagnetic radiation is $3.3\times 10^{-29}$ kg m/sec, then frequency of associated wave is:

  1. $3.0\times 10^3Hz$

  2. $6.0\times 10^3Hz$

  3. $7.5\times 10^5Hz$

  4. $1.5\times 10^{13}Hz$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p=\dfrac {hv}{c}$
$\Rightarrow v=\dfrac {cp}{h}=\dfrac {(3\times 20^8)\times (3\cdot 3\times 10{-29})}{6\cdot 6\times 10^{-34}}$


$=1\cdot 5\times 10^{13}Hz.$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Assume that a neutron breaks into a proton and an electron. The energy released suring this process is (mass of neutron $= 1.6725 \times 10^{-27} kg,$
mass od proton $= 1.6725 \times 10^{-27} kg,$ mass of electron$= 9 \times 10^{-31}kg)$

  1. 0.73 MeV

  2. 7.10 MeV

  3. 6.30 MeV

  4. 5.4 MeV
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\triangle m = (m _P + m _e) -m _n$

        $= 9 \times 10^{-31} kg $
 $(mc^2) \space energy \space released = (9 \times 10^{-31} kg) c^2 \space joules$
$= \dfrac{9 \times 10^{-31} \times (3 \times 10^8 )^2}{1.6 \times 10^{-13}} Mev$

$= 0.73 Mev$
Hence option (A) is correct

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A monochromatic source of light is placed at a large distance $d$ from a metal surface. Photoelectrons are ejected at rate $n$, the kinetic energy being $E$. If the source is brought nearer to distance $d/2$, the rate and kinetic energy per photoelectron become nearly :

  1. 2n and 2E

  2. 4n and 4E

  3. 4n and E

  4. n and 4E
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The rate is inversely proportional to square of the distance: $n\propto \dfrac{1}{{r}^{2}}$


So the new rate will be $4n$

Kinetic energy is not related to the distance, hence it will remain same $E$.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The work function of a certain metal is $3.31 \, \times \,  10^{-19} J.$ Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 A is
$(Given\, h  = 6.62\times10^{-34} J-s, \, c= 3\times10^{-8} \, ms^{-1},\, e=  1.6 \times10^{-19} \, C)$

  1. $248 eV$

  2. $0.41 eV$

  3. $2.07 eV$

  4. $0.82 eV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work function $W _0 \, = \, 3.31 \, \times \, 10^{-19} \, J$
Wavelength of incident radiation
 $\lambda \, = \, 5000 \, \times \, 10^{-10} \, m$
$E \, = \, W _0 \, + \, KE$
(According to Einstein equation)
$\displaystyle \frac{hc}{\lambda} \, 3.31 \, \times \, 10^{-19} \, + \, KE$
$KE \, = \, - \, 3.31 \, \times \, 10^{-19} + \, \displaystyle \frac{6.62 \, \times \, 10^{-34} \, \times \, 3 \, \times \, 10^8}{5000 \, \times \, 10^{10}}$
$= \, - \, 3.31 \, \times \, 10^{-19} \, + \, 10^{-19} \, + \, \displaystyle \frac{6.62 \, \times \, 3}{5} \, \times \, 10^{-19}$
$= \, (- \, 3.31 \, \times \, 1.324 \, \times \, 3) \, \times \, 10^{-19}$
$= \, (3.972 \, - \, 3.31) \, \times \, 10^{-19} \, = \, 0.662 \, \times \, 10^{-19} \, J$
$\Rightarrow \, E \, = \, \displaystyle \frac{0.662 \, \times \, 10^{19}}{1.6 \, \times \, 10^{-19}} \, = \, 0.41 eV$

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

When sound wave is refracted from air to water, which of the following will remain unchanged?

  1. wave number

  2. wavelength

  3. wave velocity

  4. frequency
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Frequency of a wave is characterised by the source of wave, therefore it will not be affected by change in medium . 

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

Ultrasound has frequency of vibration .....

  1. between 20 Hz and 20000 Hz.

  2. below 20 Hz.

  3. above 20000 Hz.

  4. between 500 Hz and 10000 Hz.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The term "Ultrasound" refers to the sound having frequency greater than the human audible frequency. Maximum limit of the human audible sound is $20000$ Hz.

So, ultrasound has  frequency of vibration above $20000$ Hz.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

1 Hz is:

  1. 1 vibration per minute.

  2. 360 vibration per minute.

  3. 10 vibration per minute.

  4. 60 vibration per minute.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Frequency of 1 Hz implies to 1 complete vibration per second. This means, 60 complete vibrations in 60 second, or, in 1 minute.