Modern Physics - JEE Main Question- mini quiz

Modern Physics - JEE Main Question

Q.1 A beam of electromagnetic radiation of intensity $6.4 \times 10^{-5} \, \text{W/cm}^2$ is comprised of wavelength, $\lambda = 310 \, \text{nm}$. It falls normally on a metal (work function $\phi = 2 \, \text{eV}$) of surface area of $1 \, \text{cm}^2$. If one in $10^3$ photons ejects an electron, total number of electrons ejected in 1 second is $10^x$.
(hc = $1240 \, \text{eV nm}$, $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$), then $x$ is _______.

Solution:

Given data:

• Intensity, $I = 6.4 \times 10^{-5} \, \text{W/cm}^2 = 6.4 \times 10^{-3} \, \text{W/m}^2$
• Wavelength, $\lambda = 310 \, \text{nm} = 310 \times 10^{-9} \, \text{m}$
• Work function, $\phi = 2 \, \text{eV}$
• Area, $A = 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2$
• Planck's constant × speed of light, $hc = 1240 \, \text{eV nm} = 1240 \times 1.6 \times 10^{-19} \, \text{J}$
• $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$

Step 1: Energy of one photon
$E = \frac{hc}{\lambda} = \frac{1240 \times 1.6 \times 10^{-19}}{310 \times 10^{-9}}$
$E \approx 6.4 \times 10^{-19} \, \text{J}$

Step 2: Number of photons falling per second
$N = \frac{\text{Intensity} \times \text{Area}}{\text{Energy of one photon}}$
$N = \frac{6.4 \times 10^{-3} \times 10^{-4}}{6.4 \times 10^{-19}}$
$N \approx 10^{12} \, \text{photons/s}$

Step 3: Number of electrons ejected per second
Since only one in $10^3$ photons ejects an electron:
$\text{Electrons ejected} = \frac{N}{10^3} = \frac{10^{12}}{10^3} = 10^9$

Therefore, $x = 9$.

Q.2 The time period of revolution of an electron in its ground state orbit in a hydrogen atom is $1.6 \times 10^{-16} \, \text{s}$. The frequency of revolution of the electron in its first excited state (in $\text{s}^{-1}$) is:

1. $6.2 \times 10^{15}$
2. $5.6 \times 10^{15}$
3. $7.8 \times 10^{14}$
4. $1.6 \times 10^{15}$

Q3.When a photon of energy $4.0 \, \text{eV}$ strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_A \, \text{eV}$ and de-Broglie wavelength $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy $4.50 \, \text{eV}$ is $T_B = (T_A - 1.5) \, \text{eV}$. If the de-Broglie wavelength of these photoelectrons $\lambda_B = 2 \lambda_A$, then the work function of metal B is:

1. 3 eV
2. 2 eV
3. 4 eV
4. 1.5 eV

Solution:

The energy of a photon is related to the maximum kinetic energy and work function as:
$E = T + \phi$, where $E$ is the photon energy, $T$ is the maximum kinetic energy, and $\phi$ is the work function.

For metal A:
$4.0 = T_A + \phi_A$
$\Rightarrow \phi_A = 4.0 - T_A$

For metal B:
$4.5 = T_B + \phi_B$
$T_B = T_A - 1.5$
$\Rightarrow 4.5 = (T_A - 1.5) + \phi_B$
$\Rightarrow \phi_B = 4.5 - (T_A - 1.5)$
$\phi_B = 6.0 - T_A$

From the de-Broglie wavelength relation, we know:
$\lambda = \frac{h}{\sqrt{2mT}}$
$\frac{\lambda_B}{\lambda_A} = \sqrt{\frac{T_A}{T_B}}$
Given, $\lambda_B = 2\lambda_A$:
$2 = \sqrt{\frac{T_A}{T_B}}$
Squaring both sides:
$4 = \frac{T_A}{T_B}$
$T_B = \frac{T_A}{4}$

Substitute $T_B = \frac{T_A}{4}$ into $T_B = T_A - 1.5$:
$\frac{T_A}{4} = T_A - 1.5$
Multiply through by 4:
$T_A = 4T_A - 6$
$3T_A = 6 \Rightarrow T_A = 2 \, \text{eV}$

Substitute $T_A = 2 \, \text{eV}$ into $\phi_B = 6.0 - T_A$:
$\phi_B = 6.0 - 2.0 = 4.0 \, \text{eV}$

Therefore, the correct answer is: (3) 4 eV.