I calculated the energy of the ground and first four excited states!! What is th

Question

I calculated the energy of the ground and first four excited states!!

What is the energy of the ground state? -8.18 eV

What is the energy of the first excited state? -2.05 eV

What is the energy of the second excited state? -0.91 eV

What is the energy of the third excited state? -0.51 eV

What is the energy of the fourth excited state? -0.33 eV

Then it asked to calculate the longest and shortest wavelength lines.

Calculate the longest-wavelength (alpha) lines in the Balmer series for this atom. 1090.50 nm

I CAN'T GET THE SHORT-WAVELENGTH SERIES!!! I have done hc/(E infinity - E2) =6048.8 nm

It says I am still wrong, Please help and show specifically what you did!! THANKS!!

Explanation / Answer

apply the formula..

1/=Rh(1/n1^2 - 1/n2^2)...........for balmer n1=2...........(Rh=rydberg's const).....;again for to be minimum... 1/ should be max....so 1/n2 should be 0 so n2 should be ............

min= 4/Rh........put value of Rh and solve...

answer= 367 nm

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