The energy of a photon of sodium light (lambd | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.To express the energy in $eV$, we use the relation $E(eV) = \frac{12400}{\

Vedclass · Accepted Answer

$2.1$

Vedclass · Answer

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.To express the energy in $eV$, we use the relation $E(eV) = \frac{12400}{\lambda(\mathring{A})}$.Given $\lambda = 589 	ext{ nm} = 5890 \mathring{A}$.Substituting the value of $\lambda$ into the formula:$E = \frac{12400}{5890} \approx 2.1 	ext{ eV}$.Thus, the minimum energy required to create an electron-hole pair is $2.1 	ext{ eV}$.

The energy of a photon of sodium light $(\lambda = 589 \text{ nm})$ is equal to the band gap of a semiconductor. Find the minimum energy $E$ in $eV$ required to create an electron-hole pair.

Similar Questions

The resultant voltage gain in a series combination of two or more amplifiers is .......

Find the value of current $I$ in the following circuits.

Select the correct statements from the following:

In which of the following circuits is the junction reverse biased?

The device that can act as a complete electronic circuit is

Vedclass Test Series

Exam Paper Generator

Online Exam Module