QJEEM24Ph829f: JEE Main 2024 Physics - Class 12 JEEM24Ph829f Solution: The acceptor level of a p-type semiconductor is 6   eV .…

The energy required to create a hole in a p-type semiconductor can be directly related to the acceptor level because this energy level represents the minimum energy required to excite an electron from the valence band into the acceptor level, effectively creating a hole. The acceptor level is given as $$6\text{eV}$$.

To find the maximum wavelength of light that can excite an electron into this level, we use the equation that relates the energy ($$E$$) of a photon to its wavelength ($$\lambda$$):

$$E=\frac{hc}{\lambda }$$

Where:

• $$E$$ is the energy in electronvolts (eV),
• $$h$$ is Planck's constant,
• $$c$$ is the speed of light, and
• $$\lambda$$ is the wavelength in nanometers (nm).

The product of $$hc$$ is given as $$1242\text{eV nm}$$, allowing us to solve for $$\lambda$$ directly:

$$\lambda =\frac{hc}{E}$$

Substituting the given values:

$$\lambda =\frac{1242\text{eV nm}}{6\text{eV}}=207\text{nm}$$

Thus, the maximum wavelength of light that can create a hole in the semiconductor by exciting an electron into the acceptor level is $$207\text{nm}$$. Therefore, the correct answer is Option D: 207 nm.