Atomic Spectra - GRE Subject Test: Physics

Card 0 of 2

Question

What color will be emitted in the to transition in the Hydrogen Balmer series?

Answer

$\frac{1}{\lambda}=R(\frac{1}{m^2}-\frac{1}{n^2})$ .

Here, $\lambda$ is the wavelength, $R (1.097 *10^7 m^{-1})$ is the Rydberg constant, for the Balmer series, and refers to the upper atomic level (this is in this case). Now, we just need to plug everything in, and solve for wavelength.

$\frac{1}{\lambda}=(1.097;x;10^7)(\frac{1}{2^2}-\frac{1}{5^2}).$

$\frac{1}{\lambda}=(1.097;x;10^7)(\frac{1}{2^2}-\frac{1}{5^2}).$

$\frac{1}{\lambda}=(1.097;x;10^7)(\frac{21}{100})=2303700m^{-1}.$

By inverting, $\lambda=4.34;x;10^{-7}m=434nm.$ This is in the violet regime of the visible spectrum.

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Question

What is the energy of the photon emitted when a Hydrogen atom makes a transition from the to the atomic energy level?

Answer

$E=-13.6eV(\frac{1}{n_f^2}-\frac{1}{n_i^2})$ , where refers to the final and initial energy levels, respectively.

$E=-13.6eV(\frac{1}{2^2}-\frac{1}{4^2})=-13.6eV\frac{3}{16}=-2.55eV.$

However, since the question asked what is the energy of the photon, it is the absolute value of the energy calculated above because of conservation of energy. The photon carries away the energy lost by the atom.

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