JEE Main Dual Nature of Radiation and Matter PYQ

1.(JEE Main 2024 (Online) 1st February Evening Shift )

Monochromatic light of frequency $6 \times 10^{14} Hz$ is produced by a laser. The power emitted is $2 \times 10^{- 3} IN$ .

How many photons per second on an average, are emitted by the source ?

(Given $h = 6.63 \times 10^{- 34} Js$ )

A.5 ⨯10¹⁵

B.7 ⨯10¹⁶

C.6 ⨯10¹⁵

D.9 ⨯10¹⁸

Correct option is (A)

To find out the number of photons emitted per second by the laser, we can use the relationship between the energy of a single photon, the total energy emitted per second (power), and the number of photons emitted per second. The energy $AND$ of a single photon is given by Planck's equation:

$AND = h f$

where:

$h$ is Planck's constant ( $6.63 \times 10^{- 34} Js$ ), and
$f$ is the frequency of the light ( $6 \times 10^{14} Hz$ ).

Let's first calculate the energy of one photon:

$AND = (6.63 \times 10^{- 34} Js) \times (6 \times 10^{14} Hz)$

$AND = 3.978 \times 10^{- 19} J$

The power ( $P$ ) emitted by the laser is the total energy emitted per second,

$P = {AND}_{total per second} = 2 \times 10^{- 3} IN = 2 \times 10^{- 3} J / s$

The number of photons ( $N$ ) emitted per second can be found by dividing the total energy emitted per second by the energy of one photon:

$N = \frac{P}{AND}$

Substitute the values we have:

$N = \frac{2 \times 10^{- 3} J / s}{3.978 \times 10^{- 19} J}$

$N = \frac{2 \times 10^{- 3}}{3.978 \times 10^{- 19}}$

$N = 5.03 \times 10^{15} photons per second$

The number of photons emitted per second is approximately $5 \times 10^{15}$ . Therefore, the correct answer, rounded to one significant figure, is:

Option A: $5 \times 10^{15}$

2.(JEE Main 2024 (Online) 1st February Evening Shift )

Conductivity of a photodiode starts changing only if the wavelength of incident light is less than $660 nm$ . The band gap of photodiode is found to be $(\frac{X}{8}) eV$ . The value of $X$ is :

(Given, $h = 6.6 \times 10^{- 34} Js, e = 1.6 \times 10^{- 19} C$ )

A.11

B.13

C.15

D.21

Correct option is (C)

To find the value of $X$ in the band gap $(\frac{X}{8}) eV$ , we need to understand the relationship between the wavelength of light that can result in changes in the conductivity of a photodiode and the photodiode's band gap energy.

The band gap energy ( $E_{g}$ ) of a material is the minimum energy required for an electron to transition from the valence band to the conduction band, thus creating a hole-electron pair and allowing conductivity to occur. When light with a certain wavelength ( $λ$ ) is incident upon the photodiode, if the energy of the photons is greater than or equal to the band gap energy of the photodiode, then the photons can excite electrons and change the conductivity of the photodiode.

The energy of a photon ( $E_{p h o t o n}$ ) is given by the equation:

$E_{p h o t o n} = \frac{h c}{λ}$

where:

$h$ is Planck's constant, which is given as $6.6 \times 10^{- 34}$ J·s.
$c$ is the speed of light in a vacuum ( $\approx 3 \times 10^{8}$ m/s).
$λ$ is the wavelength of the incident light.

Then the band gap energy in terms of electron volts is found by converting the energy from joules to electron volts (eV) using the charge of an electron $e$ ( $1.6 \times 10^{- 19}$ C):

$E_{g} = \frac{h c}{λ e}$

Given that the photodiode starts conducting when the wavelength of light is less than $660 nm$ , we should use this wavelength as the threshold wavelength $λ_{t h r e s h o l d}$ :

$E_{g} = \frac{h c}{λ_{t h r e s h o l d} \times e}$

$E_{g} = \frac{6.6 \times 10^{- 34} J·s \cdot 3 \times 10^{8} m/s}{660 \times 10^{- 9} m \times 1.6 \times 10^{- 19} C}$

$E_{g} = \frac{6.6 \times 3 \times 10^{- 26}}{660 \times 1.6} eV$

$E_{g} = \frac{19.8}{660 \times 1.6} eV$

$E_{g} = \frac{19.8}{1056} eV$

$E_{g} = \frac{18.75}{1000} eV$

$E_{g} = 1.875 eV$

Now, we have to compare this value to $(\frac{X}{8}) eV$ to find the value of $X$ :

$1.875 = \frac{X}{8}$

$X = 1.875 \times 8$

$X = 15$

Therefore, the value of $X$ is 15. The correct answer is Option C.

3.(JEE Main 2024 (Online) 30th January Evening Shift )

If the total energy transferred to a surface in time $t$ is $6.48 \times 10^{5} J$ , then the magnitude of the total momentum delivered to this surface for complete absorption will be:

A. $2.16 \times 10^{- 3} kg m / s$

B. $2.46 \times 10^{- 3} kg m / s$

C. $1.58 \times 10^{- 3} kg m / s$

D. $4.32 \times 10^{- 3} kg m / s$

Correct option is (A)

$p = \frac{E}{C} = \frac{6.48 \times 10^{5}}{3 \times 10^{8}} = 2.16 \times 10^{- 3}$

4.(JEE Main 2024 (Online) 29th January Evening Shift )

Two sources of light emit with a power of $200 W$ . The ratio of number of photons of visible light emitted by each source having wavelengths $300 nm$ and $500 nm$ respectively, will be :

A.5 : 3

B.3 : 5

C.1 : 5

D.1 ; 3

Correct option is (B)

$\begin{aligned} n_{1} \times \frac{hc}{λ_{1}} = 200 \\ n_{2} \times \frac{hc}{λ_{2}} = 200 \\ \frac{n_{1}}{n_{2}} = \frac{λ_{1}}{λ_{2}} = \frac{300}{500} \\ \frac{n_{1}}{n_{2}} = \frac{3}{5} \end{aligned}$