The figure below depicts the first four energy levels in a hydrogen atom. The th

Question

The figure below depicts the first four energy levels in a hydrogen atom. The three transitions shown as arrows emit ultraviolet light and occur at wavelengths of 121.566 nm, 102.583 nm, and 97.524 nm. respectively. Planck's constant is 6.62607 times 10^-34 J s and the speed of light is 2.99792 times 10^8 m/s. Find the frequency of light that would be emitted in a transition from the state labeled as n = 4 to the state labeled as n = 2. Frequency = Hz Find the wavelength of light that would be emitted when a hydrogen atom undergoes the transition from the state labeled as n = 3 to the state labeled as n = 2 nm. Express your answer in nm. Wavelength = nm

Explanation / Answer

We know that the wavelength and enegy are inversely proportional . Thus the shortest wavelength corresponds to largest energy transition and longest wave length corresponds to smallest energy tansition.

Thus the wavelength 97.524 nm corresponds to 4--> 1 transition,

102.583   nm corresponds to 3--> 1 transition, and

121.566  nm corresponds to 2--> 1 transition,

From borhr's theorey we know

E(n2) -E(n1) = hc/lambda

Thus E4 - E1 = 6.62607 x 10-34 x 2.99792x108  / 97.524x10-9

= 2.03687 x10-18 J

similarly E2 - E1= 6.62607 x 10-34 x 2.99792x108  / 121.566x10-9

= 1.6340 x10-18 J

similarly E3 - E1= 6.62607 x 10-34 x 2.99792x108  / 102.583x10-9

= 1.9364 x10-18 J

Part a)

We can write the E4 -E2 as (E4-E1) - (E2 -E1)

Thus E4 -E2 = 2.03687 x10-18 J -1.6340 x10-18 J

= 4.0287 x10-19 J

Frequency = energy / h

= 4.0287 x10-19J/6.62607 x 10-34

= 6.08x1014 HZ

part b

similarly E3 -E2 can be taken (E3-E1) - (E2-E1)

Thus E3 -E2 = 1.9364 x10-18 J-1.6340 x10-18 J

= 3.024x10-19 J

The wavelength corresponding = hc/E

=  6.62607 x 10-34 x 2.99792x108 / 3.024x10-19 J

= 6.5689 x10-7 m

= 658.9 x10-9 m m

Thus the qvelength is 658.9 nm

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