Electrons accelerated by a potential difference of 12.87 V pass through a gas of hydrogen atoms at room temperature. Calculate the wavelength of light emitted with the longest possible wavelength.

I don't know what equation to use. Any help would be greatly appreciated.

You know the KE of the electrons. They can pass this quantized energy to the hydrogen atom. Set this equal to 13.6v

http://musr.physics.ubc.ca/~jess/hr/skept/QM1D/node4.html

or
12.87>13.6 (1/n^2 - 1/m^2) What n, m will work? The longest wavelength will occur at the least energy, or n is very large, and m is infinity. So the question bothers me, if it read calculate the shortest wavelength, then energy would be as close to 12.87 as possible, the highest frequency (low wavelength). So if it is that, then n=1, m=very large will work. Then
12.87=plancks constant(c/lambda) and you can calculate lambda. But again, this is the shortest wavelength.
Check my thinking.

To calculate the wavelength of light emitted with the longest possible wavelength, we can use the equation:

λ = hc/E

Where:
λ is the wavelength of light emitted,
h is the Planck's constant (6.626 x 10^-34 J.s),
c is the speed of light in vacuum (3 x 10^8 m/s),
and E is the energy of the emitted photon.

In this case, we need to find the energy of the emitted photon.

The energy of a photon can be calculated using the equation:

E = qV

Where:
E is the energy of the photon,
q is the charge of the electron (1.6 x 10^-19 C),
and V is the potential difference (12.87 V).

Substituting the values, we have:

E = (1.6 x 10^-19 C) * (12.87 V)
E = 2.052 x 10^-18 J

Now, we can substitute the value of E back into the first equation to find the wavelength:

λ = (6.626 x 10^-34 J.s) * (3 x 10^8 m/s) / (2.052 x 10^-18 J)

Calculating this will give you the wavelength of light emitted with the longest possible wavelength.