How do I derive the Rydberg constant from the Bohr Formula from the plank-einstein relation Ephoton=hf?

Question

anonymous · Answer

I currently have the following 

E = Einitial - Efinal 
Ephoton = hc / lambda

aravindg · Answer

hmm . ..nic qn w8 lemme think ovr it

aravindg · Answer

seems u need to equate lambda

anonymous · Answer

ther is an amguity in rydberg const...it can mean the const coming in hydrogen spectra or in the energy of orbits equation..

shayaan_mustafa · Answer

Hello may I help here?

fretje · Answer

@shayaan_Mustafa yes i think you may

ujjwal · Answer

according to bohr's formula energy released, when an eletron makes a transition from a higher energy level n2 to lower energy level n1, is given by$$E=R \left( 1/n1^{2} -1/n2^{2}\right)$$ where r is rydberg constant..
At the same time the energy of photon released during the process is given by E= hf
Equalise the two values of energy.. if values of n1, n2, and f is known R can be obtained easily..

ujjwal · Answer

*R