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

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

  1. 3 V

  2. 3 eV

  3. 6 eV

  4. 9 eV
Reveal answer Fill a bubble to check yourself
B Correct answer
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).

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

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

  1. Doubled

  2. Halved

  3. Quadrupled

  4. More than twice the previous value
Reveal answer Fill a bubble to check yourself
D Correct answer
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).

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

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 $hc= 2 \times 10^{-25} Jm$]

  1. $3.1 eV$

  2. $2.48 eV$

  3. $0.62 eV$

  4. $5. 58 eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
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).

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

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)

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

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

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

  4. $2.98\times 10^{-15}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
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).

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

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:

  1. 2.48eV ,0.18eV

  2. 0.18eV,Zero

  3. 2.30eV,Zero

  4. 0.18eV,0.18eV
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

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

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

  1. 1

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

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

  4. $\displaystyle \sqrt{\frac{m _e}{2m _\alpha}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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.

  1. thermodynamics

  2. quantum mechanics

  3. classical electrodynamics

  4. special relativity

  5. general relativity
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\Delta x.\Delta p\geq\dfrac{\hbar}{2}$

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

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

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

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

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

  4. $\displaystyle\lambda^{'}=\frac{\lambda _{0}}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $\displaystyle \frac{\lambda _{0}}{{\lambda}'}=\sqrt{\frac{{E}'}{E}}=\sqrt{\frac{2E}{E}}=\sqrt{2}$
$\displaystyle \therefore  {\lambda}'= \frac{\lambda _{0}}{\sqrt{2}}$

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

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

  1. $300 p'$

  2. $500 p'$

  3. $400 p'$

  4. $100 p'$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $p = 500 p'$
Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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 $9v _{0}/4$?

  1. $3\ eV$

  2. $4.5\ eV$

  3. $6\ eV$

  4. $9\ eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
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), $\phi _{0} = hv _{0} - 1\ eV$
$= 4\ eV - 1\ eV = 3\ eV$
$\therefore (KE) _{mas} = h\times \dfrac {9v _{0}}{4} - 3\ eV$
$= \dfrac {9}{4} (4\ eV) - 3\ eV = 6\ eV$.