Tag: physics

Questions Related to physics

The cut-off voltage in a photoelectric experiment is 3V. Then the maximum KE of photo-electrons emitted is

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 3 V

  2. 3 eV

  3. 6 eV

  4. 9 eV

Show answer
Correct Option: B
Explanation:

Cut off voltage is the minimum voltage applied across the plates that even the electrons ejected with minimum kinetic energy could not reach the other plate.
So, 
From the definition,
Cut off voltage $= 3V$
Work done on the charge $=$ Kinetic energy of the photons
$\Rightarrow 3V \times 1e = K. E.$
$\Rightarrow K. E. = 3eV$

So, the answer is option (B).

If the frequency of light incident on a photosensitive metal plate is doubled, then the KE of photoelectrons will be

physics quantum physics photons concept of photon photons and photoelectric effect

  1. Doubled

  2. Halved

  3. Quadrupled

  4. More than twice the previous value

Show answer
Correct Option: D
Explanation:

$\ KE\ =\ h(V-V _{0})$
$\ V^{1}\ =\ 2V$
$\ KE^{1}\ =\ h(2V-V _{0})$
$\ KE^{1}\ =\ 2h(V-V _{0}) +\ hV _{0}$
$\ KE^{1}\ =\ 2KE+hV _{0}$

So, the answer is option (D).

The photo-electric threshold wavelength for a metal is $5000 A^{0}$. Light of wavelength $4000 A^{0}$ is incident on it. The maximum KE of photo-electrons emitted is [given $hc= 2 \times 10^{-25} Jm$]

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $3.1 eV$

  2. $2.48 eV$

  3. $0.62 eV$

  4. $5. 58 eV$

Show answer
Correct Option: C
Explanation:

$K _{max}\ =\ \dfrac{hc}{\lambda}-\dfrac{hc}{\lambda _{0}}$

            $=\dfrac{20\times 10^{-26}}{4\times 10^{-7}}-\dfrac{20\times 10^{-26}}{5\times 10^{-7}}$

           $=\ 20\times 10^{-19}\left ( \dfrac{1}{4} -\dfrac{1}{5}\right )\ J$

          $=10^{-19}J=\dfrac{10^{-19}}{1.6\times 10^{-19}}eV=0.62eV$

So, the answer is option (C).

An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.40$A^{0}$. Then the maximum energy of the photon of the emitted radiation. 

($h = 6.63 \times10^{-34}Js$ and $c= 3 \times10^{8}$ m/s)

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $4.9725\times 10^{-15}J$

  2. $6.28\times 10^{-15}J$

  3. $3.15\times 10^{-15}J$

  4. $2.98\times 10^{-15}J$

Show answer
Correct Option: A
Explanation:

$\lambda _{cutoff} =\ \dfrac{hc}{E}$

$E _{max}=\ \dfrac{20\times 10^{-26}}{0.4\times 10^{-10}}$

$=\ 4.9725\times 10^{-15}\ J$

So, the answer is option (A).

The work function of a certian metal is 2.3 eV .If light of wave number $2\times10^{6}m^{-1}$ falls on it,the kinetic energies of fastest and slowest ejected electorn will be respectively:

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 2.48eV ,0.18eV

  2. 0.18eV,Zero

  3. 2.30eV,Zero

  4. 0.18eV,0.18eV

Show answer
Correct Option: B
Explanation:

$\phi _{ _0}=2.3eV$ 


$\lambda^{-1}=2\times10^6m^{-1}$


Some electrons, although get loosened due to the incident rays, remain stationary.
$\therefore KE _{min}=0$ always

$KE _{max}=TE-\phi _{ _0}$  where TE is the energy of the rays.

$TE=hc\lambda^{-1}$  in Joules

$TE=\dfrac{hc\lambda^{-1}}e$ $in$ $eV$ where e is the charge of one electron

$\therefore TE=\dfrac{6.63\times10^{-34}\times{3\times10^8}\times2\times 10^6}{1.6\times10^{-19}}=2.48eV$

$\therefore KE _{max}=2.48-2.3=0.18eV$

The ratio of momentum  of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of 100 V is

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 1

  2. $\displaystyle \sqrt{\frac{2m _e}{m _\alpha}}$

  3. $\displaystyle \sqrt{\frac{m _e}{m _\alpha}}$

  4. $\displaystyle \sqrt{\frac{m _e}{2m _\alpha}}$

Show answer
Correct Option: C
Explanation:

Since both particles are accelerated by same potential difference, their kinetic energies are equal.

Thus $\dfrac{1}{2}m _{e}v _e^2=\dfrac{1}{2}m _{\alpha}v _{\alpha}^2$
$\implies \dfrac{v _e}{v _{\alpha}}=\sqrt{\dfrac{m _{\alpha}}{m _e}}$
Thus the ratio of their momenta=$\dfrac{m _ev _e}{m _{\alpha}v _{\alpha}}=\sqrt{\dfrac{m _e}{m _{\alpha}}}$
Hence correct answer is option C.

Which one of the following branch of physics deals the impossibility of making simultaneous, arbitrarily precise measurements of the momentum and the position of an electron.

physics quantum physics photons concept of photon photons and photoelectric effect

  1. thermodynamics

  2. quantum mechanics

  3. classical electrodynamics

  4. special relativity

  5. general relativity

Show answer
Correct Option: B
Explanation:

Heisenberg's Uncertainty Principle, as a part of quantum mechanics, theorizes that the position and momentum of an electron cannot be measured precisely at the same time, that is, there is a minimum associated natural error in measurement of the two simultaneously. 

The error in measurement of the two is quantified as $\Delta x.\Delta p\geq\dfrac{\hbar}{2}$

In photoelectric effect, initially when energy of electrons emitted is $E _{0}$, de-Broglie wavelength associated with them is $\lambda _{0}$. Now, energy is doubled then associated de-Broglie wavelength $\lambda^{'}$ is

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\displaystyle\lambda^{'}=\frac{\lambda _{0}}{\sqrt{2}}$

  2. $\displaystyle\lambda^{'}=\sqrt{2}\lambda _{0}$

  3. $\displaystyle\lambda^{'}=\lambda _{0}$

  4. $\displaystyle\lambda^{'}=\frac{\lambda _{0}}{2}$

Show answer
Correct Option: A
Explanation:

de-Brogile wavelength is given by
$\displaystyle\lambda =\frac{h}{p}$, where h= Planck's constant and p= momentum
Also energy (E) and momentum are related as 
$\displaystyle E =\frac{p^{2}}{2m}$
$\displaystyle \therefore p=\sqrt{2mE}$
$\displaystyle \therefore \lambda =\frac{h}{\sqrt{2mE}}\times \frac{1}{\sqrt{E}}$ as h and m are constants
Hence, $\displaystyle \frac{\lambda _{0}}{{\lambda}'}=\sqrt{\frac{{E}'}{E}}=\sqrt{\frac{2E}{E}}=\sqrt{2}$
$\displaystyle \therefore  {\lambda}'= \frac{\lambda _{0}}{\sqrt{2}}$

When the momentum of a photon is changed by an amount $p'$ then the corresponding change in the de-Broglie wavelength is found to be $0.20$%. Then, the original momentum of the photon was

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $300 p'$

  2. $500 p'$

  3. $400 p'$

  4. $100 p'$

Show answer
Correct Option: B
Explanation:
As, we know de-Broglie wavelength,
$\lambda = \dfrac{h}{p}$
$\therefore \lambda \propto \dfrac{1}{p}$
$\Rightarrow \dfrac{\Delta p}{p} = - \dfrac{\Delta \lambda}{\lambda} \therefore \left| \dfrac{\Delta p}{p} \right | = \left| \dfrac{\Delta \lambda}{\lambda} \right |$
$\Rightarrow \dfrac{p'}{p} = \dfrac{0.20}{100} = \dfrac{1}{500}$
or, $p = 500 p'$

In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is $1eV$ for incoming radiation of frequency $v _{0}$ and $3eV$ for incoming radiation of frequency $3v _{0}/2$. What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency $9v _{0}/4$?

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $3\ eV$

  2. $4.5\ eV$

  3. $6\ eV$

  4. $9\ eV$

Show answer
Correct Option: C
Explanation:

$(KE) _{max} = hv - \phi _{0}$
So, $1\ eV = hv _{0} - \phi _{0} .... (i)$
and $3\ eV = \dfrac {hv _{0}}{2} - \phi _{0} .... (ii)$
$\Rightarrow 3\ eV - 1\ eV = \dfrac {hv _{0}}{2}$
or $hv _{0} = 4\ eV$
From Eq. (i), $\phi _{0} = hv _{0} - 1\ eV$
$= 4\ eV - 1\ eV = 3\ eV$
$\therefore (KE) _{mas} = h\times \dfrac {9v _{0}}{4} - 3\ eV$
$= \dfrac {9}{4} (4\ eV) - 3\ eV = 6\ eV$.