Answer to Question #106201 in Chemistry for shaniza

Question

What is the energy possessed by a radiation of wavenumber 1250 cm–1

Answer 1

Solution:

n (wavenumber) = 1250 cm^–1.

1)  

n (wavenumber) = 1 / λ;

λ (wavelength) = 1 / n = 1 / (1250 cm^-1) = 0.0008 cm.

2) c = 3·10¹⁰ cm/s.

ν (frequency) = c / λ;

ν (frequency) = (3·10¹⁰ cm/s) / (0.0008 cm) = 3.75·10¹³ s^-1.

3) h = 6,63⋅10⁻³⁴ kg·m²·s⁻¹

The Einstein equation, E = h * ν , will give the energy associated with one photon. 

E (energy) = h * ν;

E (energy) = (6,63⋅10⁻³⁴ kg·m²·s⁻¹) * (3.75·10¹³ s^-1) = 2.486·10^-20 kg·m²·s⁻² = 2.486·10^-20 J

E (energy) = 2.486·10^-20 J per photon.

We also can to calculate the total energy (in Joules) associated with 1 mole of photons. For it We need to multiply the energy obtained by Avogadro’s number. 

Na = 6.02⋅10²³ photons / mole

E (energy) = E * N_a;

E (energy) = (2.486·10^-20 J / photon) * (6.02⋅10²³ photons / mole) = 14965.72 J/mole = 15 kJ / mole;

E (energy) = 14965.72 J per mole of photons.

Answer:

E (energy) = 2.486·10^-20 J per photon.

E (energy) = 14965.72 J per mole of photons.