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 ) for blue light of a similar wavelength to be emitted?

n = 3

Explanation:

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

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

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

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.

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.