Some books give the equation for the emission of a photon in hydrogen as follows: 1/(lamda)=R(1/n1^2 - 1/n2^2) where n1 is the quantum number in the lower state and n2 is the quantum number in the higher energy state, and R= 1.097x10^7 m. Calculate the energy of a photon emitted when an electron jumps from n=5 to the n=2 level.
I have this part done, however I'm not sure if i should multipy 2302700 or 434.1 in the equation E=hv/(lamba)
Some books give the equation for the emission of a photon in hydrogen as lower states follows: 1 R( 1/n 1 /n22) where ni is the quantum number in the and n, is the quantum number in the higher energy state, and R 1.097 x 107 m to Calculate the energy of a photon emitted when an electron jumps from the n 5 the n 52 level. t
