JEE Main Dual Nature of Radiation and Matter PYQ

51. (AIEEE 2011 )

This question has Statement - $1$ and Statement - $2$ . Of the four choices given after the statements, choose the one that best describes the two statements.

Statement - $1$ : A metallic surface is irradiated by a monochromatic light of frequency $v > v_{0}$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $K_{max}$ and $V_{0}$ respectively. If the frequency incident on the surface is doubled, both the $K_{max}$ anmd $V_{0}$ are also doubled.
Statement - $2$ : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

(A) Statement - $1$ is true, Statement - $2$ is true, Statement - $2$ is the correct explanation of Statement - $1$ .

(B) Statement - $1$ is true, Statement - $2$ is true, Statement - $2$ is not the correct explanation of Statement - $1$ .

(C) Statement - $1$ is false, Statement - $2$ is true.

(D) Statement - $1$ is true, Statement - $2$ is false.

Correct answer is option C

By Einstein photoelectric equation,

$K_{max} = e V_{0} = h v - h v_{0}$

When $v$ is doubled, $K_{max}$ and $V_{0}$ become more than double.

52. (AIEEE 2010 )

Statement - $1$ : When ultraviolet light is incident on a photocell, its stopping potential is $V_{0}$ and the maximum kinetic energy of the photoelectrons is $K_{max}$ . When the ultraviolet light is replaced by $X$ -rays, both $V_{0}$ and $K_{max}$ increase.

Statement - $2$ : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

(A) Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is the correct explanation of Statement - $1$

(B) Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is not the correct explanation of Statement - $1$

(C) Statement - $1$ is is false, Statement - $2$ is true

(D) Statement - $1$ is is true, Statement - $2$ is false

Correct answer is option B

Statement 1 is true. The energy of an incident photon (from the ultraviolet light or X-rays) on a photocell is given by Planck's equation, $E = h ν$ , where $h$ is Planck's constant and $ν$ is the frequency of the light. X-rays have a higher frequency than ultraviolet light, so they deliver more energy to the photoelectrons. This results in a higher stopping potential ( $V_{0}$ ) and maximum kinetic energy ( $K_{max}$ ) for the photoelectrons.

Statement 2 is also true. However, while the speeds (and hence kinetic energies) of photoelectrons do vary, this variation is not because of a range of frequencies in the incident light. Rather, it's due to the interaction of the incident photons with electrons at different energy levels in the metal. A single frequency of light can produce photoelectrons with a range of speeds because the electrons they encounter can have a variety of binding energies.

53. (AIEEE 2009 )

The surface of a metal is illuminated with the light of $400$ $n m .$ The kinetic energy of the ejected photoelectrons was found to be $1.68$ $e V .$ The work function of the metal is : $(h c = 1240 e V . n m)$

(A) $1.41$ $e V$

(B) $1.51$ $e V$

(C) $1.68$ $e V$

(D) $3.09$ $e V$

Correct answer is option A

The photoelectric effect equation, which relates the energy of the incident light to the kinetic energy of the ejected photoelectrons and the work function of the metal, is given by:

$E = K_{max} + W$ ,

where

$E$ is the energy of the incident light,

$K_{max}$ is the maximum kinetic energy of the photoelectrons, and

$W$ is the work function of the metal.

The energy of the incident light can be calculated using the formula $E = \frac{h c}{λ}$ , where $h$ is Planck's constant, $c$ is the speed of light, and $λ$ is the wavelength of the light. However, given that $h c = 1240$ eV⋅nm, we can simplify this to $E = \frac{1240}{λ}$ .

Substituting the given values into this equation, we have:

$E = \frac{1240}{400} = 3.1$ eV.

We can then substitute these values into the photoelectric effect equation:

$3.1 eV = 1.68 eV + W$ ,

which simplifies to:

$W = 3.1 eV - 1.68 eV = 1.42$ eV.

Rounding to two decimal places, the work function of the metal is therefore approximately $1.42$ eV.

Thus, Option A: $1.41$ eV is the closest to the correct answer.

54. (AIEEE 2006 )

The threshold frequency for a metallic surface corresponds to an energy of $6.2$ $e V$ and the stopping potential for a radiation incident on this surface is $5 V .$ The incident radiation lies in

(A) ultra-violet region

(B) infra-red region

(C) visible region

(D) x - ray region

Correct answer is option A

The energy of the incident radiation that causes photoelectrons to be emitted can be calculated using the stopping potential, V, via the equation

$E = e V,$

where e is the elementary charge.

The elementary charge, e, is approximately equal to $1.602 \times 10^{- 19}$ C. So the energy of the incident radiation is

$E = 1.602 \times 10^{- 19} C \times 5 V = 8.01 \times 10^{- 19} J .$

We convert this to electron volts (eV) by using the conversion factor $1 eV = 1.602 \times 10^{- 19} J .$ So

$E = \frac{8.01 \times 10^{- 19} J}{1.602 \times 10^{- 19} J/eV} = 5 eV .$

This is less than the threshold energy of $6.2 eV,$ which means that the incident radiation does not have enough energy to overcome the work function of the metal and thus cannot cause photoelectrons to be emitted.

The regions of the electromagnetic spectrum are generally classified by energy as follows:

Infrared: Less than about $1.24 eV$
Visible: Between about $1.24 eV$ and $3.1 eV$
Ultraviolet: Between about $3.1 eV$ and $124 eV$
X-rays: Greater than about $124 eV$

Therefore, the incident radiation falls in the ultraviolet region.

55. (AIEEE 2006 )

The time taken by a photoelectron to come out after the photon strikes is approximately

(A) 10 ^-4s

(B) 10 ^-10s

(C) 10 ^-16s

(D) 10 ^-1s

Correct answer is option B

Emission of photo-electron starts from the surface after incidence of photons in about 10 ^-10s