Tag: photons and photoelectric effect

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

The photoelectric cut off voltage in a certain experiment is 1.5 V. The maximum kinetic energy of photoelectrons emitted is then

  1. 2.4 eV

  2. 1.5 eV

  3. 3.1 eV

  4. 4.5 eV
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The minimum negative potential applied to the plate or anode  for which the photoelectric current just becomes zero, So, in this case, maximum K.E of an electron will be equal to stopping potential.
Here, $V _0 \, = \, 1.5 \, V,$
Maximum Kinetic energy = $eV _{0} \, = \, 1.5 \,eV$

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

Light of wavelength 0.6 mm from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 mm from a sodium lamp, the stopping potential is 1.5 V. With this data, the value of h/e is:

  1. $6\times {{10}^{-5}}\,V{{s}^{-1}} $

  2. $2\times {{10}^{-15}}\,V{{s}^{-1}} $

  3. $4\times {{10}^{-55}}\,V{{s}^{-1}} $

  4. $4\times {{10}^{-15}}\,V{{s}^{-1}} $

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In emission of electron, Potential Energy

$eV = \dfrac{hc}{\lambda} - W _0$

When light of wavelength $\lambda =0.6\,mm$and stopping potential $0.5\,V$
$0.5e=\dfrac{hc}{6\times {{10}^{-7}}}-{{W} _{0}}\ ......\ (1)$

When light of wavelength $\lambda =0.4\,mm$and stopping potential $1.5\,V$
$1.5e=\dfrac{hc}{4\times {{10}^{-7}}}-{{W} _{0}}\ ......\ (2)$
subtract equation (1) from (2) $ e=\dfrac{hc}{{{10}^{-7}}}\left[ \dfrac{1}{4}-\dfrac{1}{6} \right] $

$ \Rightarrow \dfrac{h}{e}=\dfrac{12\times {{10}^{-7}}}{3\times {{10}^{8}}}=4\times {{10}^{-15}}\,V{{s}^{-1}} $ 

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

Photons absorbed in meter are converted to heat. A source emitting $n$ photons/s of frequency $v$ is used to convert $1\ kg$ of ice of ${0}^{o}c$ to water at ${0}^{o}C$. Then, the time taken for the conversion:

  1. decreases with increasing $n$, with $v$ fixed

  2. decreases with $n$ fixed, $v$ increasing

  3. remains constant with $n$ and $v$ changing such that $nv=$constant

  4. increases when the product $nv$ increases
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The energy absorbed per second is P = n * h * v. Since the energy required to melt the ice is constant, the time taken is inversely proportional to the power P. Thus, increasing n (with v fixed) increases power and decreases time.

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

The energy od proton in a light of wavelength 6000 $\mathring { A } $ is :
$(h = 6.63 \times 10^{-34}$ $Js,$ $c = 3 \times 10^8 m/s)$

  1. $6.81 \times 10^{-19} J$

  2. $5.61 \times 10^{-19} J$

  3. $3.31 \times 10^{-19} J$

  4. $2.31 \times 10^{-19} J$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$E = {{hc} \over \alpha }$

$ = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {6000 \times {{10}^{ - 10}}}}$

$ = 3.31 \times {10^{ - 19}}J$

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

A laser beam ($\iota = 633\ nm$)has an power of $3mW$. What will be the pressure exerted on a surface by this beam if the cross sectional area is $3\ mm^{2}$ (Assume perfect reflection and normal incidence)

  1. $6.6\times 10^{-3}N/m^{2}$

  2. $6.6\times 10^{-6}N/m^{2}$

  3. $6.6\times 10^{-9}N/m^{2}$

  4. $6.6\ N/m^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pressure P = (2 * I) / c for perfect reflection. Intensity I = Power / Area = 3e-3 / 3e-6 = 1000 W/m^2. P = 2 * 1000 / 3e8 = 6.66e-6 N/m^2.

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

A metalic surface is irradiated with monochr matic light of variable wavelength. Above wavelength of $ 5000 \dot { A }  $ , no photoelectrons a emitted from the surface. With an unknown wavelength, a stopping potential of 3V is nessary to eliminate the photo current . The Known wavelength is:

  1. 2258 $ \dot { A } $

  2. $ 4133 \dot { A } $

  3. $ 3126 \dot { A } $

  4. $ 2679 \dot { A } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work function phi = 12400 / 5000 = 2.48 eV. Stopping potential 3V means max KE = 3 eV. Total energy E = 2.48 + 3 = 5.48 eV. Wavelength = 12400 / 5.48 = 2262 Angstroms. Option A is the closest.

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

Two free protons are separated by a distance of $1\mathring A $. If they are released the kinetic energy of proton when it infinite separation is

  1. $23 \times 10 ^ { 19 } j$

  2. $11.5 \times 10 ^ { - 19 } \mathrm { J }$

  3. $46 \times 10 ^ { - 19 } \mathrm { J }$

  4. $5.6 \times 10$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Potential energy U = k * q1 * q2 / r. U = 9e9 * (1.6e-19)^2 / 1e-10 = 23.04e-19 J. Since two protons share this energy, each gets half: 11.52e-19 J.

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

The maximum energy of emitted photo electrons is measured by

  1. The current they produce

  2. The potential difference they produce

  3. The largest potential difference they can traverse

  4. The speed with which they emerge
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Maximum energy of emitted photo electrons is equal to the stopping potential. Stopping potential is the largest potential difference an emitted photo electron can traverse.