Q102 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Chemistry
Calculate the energy per mole of photons of red light with a wavelength $750$ nm. [$h = 6.626 \times 10^{-34}$ Js, $C = 3 \times 10^8$ m/s.]
- Correct answerA$1.596 \times 10^5$ J
- B$2.534 \times 10^5$ J
- C$2.916 \times 10^5$ J
- D$3.138 \times 10^7$ J
Solution
The energy of a single photon is given by $E = \dfrac{hc}{\lambda}$.
The energy of one mole of photons is $E_{mole} = N_A \dfrac{hc}{\lambda}$, where $N_A$ is Avogadro's number ($6.022 \times 10^{23}$ mol$^{-1}$).
Given:
$\lambda = 750 \text{ nm} = 750 \times 10^{-9}$ m
$h = 6.626 \times 10^{-34}$ J s
$c = 3 \times 10^8$ m/s
Substituting the values into the formula:
$E_{mole} = \dfrac{6.022 \times 10^{23} \times 6.626 \times 10^{-34} \times 3 \times 10^8}{750 \times 10^{-9}}$
$E_{mole} = \dfrac{119.7 \times 10^{-3}}{750 \times 10^{-9}}$
$E_{mole} = 0.1596 \times 10^6 \text{ J mol}^{-1}$
$E_{mole} = 1.596 \times 10^5 \text{ J mol}^{-1}$
Answer: $1.596 \times 10^5$ J.
The energy of one mole of photons is $E_{mole} = N_A \dfrac{hc}{\lambda}$, where $N_A$ is Avogadro's number ($6.022 \times 10^{23}$ mol$^{-1}$).
Given:
$\lambda = 750 \text{ nm} = 750 \times 10^{-9}$ m
$h = 6.626 \times 10^{-34}$ J s
$c = 3 \times 10^8$ m/s
Substituting the values into the formula:
$E_{mole} = \dfrac{6.022 \times 10^{23} \times 6.626 \times 10^{-34} \times 3 \times 10^8}{750 \times 10^{-9}}$
$E_{mole} = \dfrac{119.7 \times 10^{-3}}{750 \times 10^{-9}}$
$E_{mole} = 0.1596 \times 10^6 \text{ J mol}^{-1}$
$E_{mole} = 1.596 \times 10^5 \text{ J mol}^{-1}$
Answer: $1.596 \times 10^5$ J.
Hence, the correct answer is option: 'A'.