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GRE Subject Test: Physics : Atomic Spectra

Study concepts, example questions & explanations for GRE Subject Test: Physics

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Example Questions

Example Question #1 : Atomic Properties

What is the energy of the photon emitted when a Hydrogen atom makes a transition from the n=4 $\displaystyle n=4$ to the n=2 $\displaystyle n=2$ atomic energy level?

Possible Answers:

2.55 eV $\displaystyle 2.55 eV$

6.8eV $\displaystyle 6.8eV$

3.4 eV $\displaystyle 3.4 eV$

13.6eV $\displaystyle 13.6eV$

5.95eV $\displaystyle 5.95eV$

Correct answer:

2.55 eV $\displaystyle 2.55 eV$

Explanation:

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

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

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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Example Question #1 : Quantum Mechanics And Atomic Physics

What color will be emitted in the n=5 $\displaystyle n=5$ to n=2 $\displaystyle n=2$ transition in the Hydrogen Balmer series?

Possible Answers:

Yellow

Violet

Green

Blue

Red

Correct answer:

Violet

Explanation:

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

Here, $\lambda$ $\displaystyle \lambda$ is the wavelength, $R (1.097 *10^7 m^{-1})$ $\displaystyle R (1.097 *10^7 m^{-1})$ is the Rydberg constant, m=2 $\displaystyle m=2$ for the Balmer series, and $\displaystyle n$ refers to the upper atomic level (this is $\displaystyle 5$ in this case). Now, we just need to plug everything in, and solve for wavelength.

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

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

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

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GRE Subject Test: Physics : Atomic Spectra

Study concepts, example questions & explanations for GRE Subject Test: Physics

All GRE Subject Test: Physics Resources

Example Questions

Example Question #1 : Atomic Properties

Example Question #1 : Quantum Mechanics And Atomic Physics

All GRE Subject Test: Physics Resources

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