Q50 of 69 Page 65

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Given:

Wavelength, λ1 = 589 nm


Wavelength, λ2 = 589.6 nm


To find frequency:


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Frequency, v1 = [3×108] / [589 × 10-9]


v1 = 5.093×1014 Hz


Frequency, v2 = [3×108] / [589.6 × 10-9]


V2 = 5.088×1014 Hz


Change in Energy, ∆E = E2-E1 = h[v2-v1]


= [6.626× 10-34] {[5.093 - 5.088] × 1014}


= [6.626× 10-34] {[5× 10-3] × 1014}


= 3.313× 10-22J


Therefore, the energy difference between two excited states is 3.31× 10-22J.


More from this chapter

All 69 →
48

In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

 49

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the Nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.

 51

The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

 52

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

λ (nm) 500 450 400


v × 10–5 (cm s–1) 2.55 4.35 5.35