© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Known Quantities
• Threshold (work function) energy $W = 6.2\,\text{eV}$.
• Stopping potential $V_0 = 5\,\text{V}$.
Step 2: Apply the Photoelectric Effect Equation
According to Einstein’s photoelectric equation,
$K_{\text{max}} = h\nu - W,$
where $K_{\text{max}}$ is the maximum kinetic energy of the emitted electrons, $h\nu$ is the energy of the incident photon, and $W$ is the work function (threshold energy).
The maximum kinetic energy of the electrons can also be expressed as:
$K_{\text{max}} = eV_0,$
where $e$ is the charge of an electron ($1.602\times10^{-19}\text{ C}$) and $V_0$ is the stopping potential.
Step 3: Compute the Photon Energy
We know:
$K_{\text{max}} = eV_0 = 5\,\text{eV}.$
Thus, the energy of the incident photon $E_{\text{photon}} = W + K_{\text{max}} = 6.2\,\text{eV} + 5\,\text{eV} = 11.2\,\text{eV}.$
Step 4: Determine the Region of the Electromagnetic Spectrum
Typical energy ranges of a few spectral regions are approximately:
Infrared (IR): less than $1.24\,\text{eV}$
Visible: about $1.24\,\text{eV}$ to $3.1\,\text{eV}$
Ultraviolet (UV): about $3.1\,\text{eV}$ to $124\,\text{eV}$
X-rays: greater than $124\,\text{eV}$
Since $11.2\,\text{eV}$ lies inside the range of ultraviolet energy, the incident radiation belongs to the ultraviolet region.
