Question

When an excited electron in a hydrogen atom falls from ????=5 to ????=2, a photon of blue light is emitted. If an excited electron in an He+ ion falls from ????=4, which energy level must it fall to (????1) for blue light of a similar wavelength to be emitted?

Answer 1

Answer:

n = 3

Explanation:

Given the formula for the transition energy of an atom with 1 electron:

$E=-13.6*Z^{2}*(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}} ) eV$

For the H transition n=5 to n=2:

$E=-13.6*(\frac{1}{4}-\frac{1}{25} ) eV=-2.856 eV$

Then we solve for nf with Z=2 (Helium)

$n_{f}=\sqrt{\frac{n_{i}^{2}*Z^{2}*13.6 eV }{2.856eV*n_{i}^{2}+13.6eV*Z^{2}} }$

$n_{f}=\sqrt{\frac{4^{2}*2^{2}*13.6 eV }{2.856eV*4^{2}+13.6eV*2^{2}} }=3$

Is near 3, actually the energy of the transitions are:

H (5⇒2) = -2.85 eV = 434 nm (Dark blue)

He (4⇒3) = -2.64 eV = 469 nm (Light blue)

I thought it was cool to see the actual colors. Included them.