Question

Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength is 40.5 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength?

Answer 1

we know, energy of electron in nth orbit in hydrogen like atom,
En = -13.6*z^2/n^2 eV

for Li+2(z = 3)

Energy of photon emited with shortest wavelength or highest energy

E = 13.6*z^2*(1/nf^2 - 1/ni^2) eV

Emax = 13.6*3^2*(1/1^2 - 1/2^2) eV

h*c/lamda_Li = 13.6*3^2*(1/1^2 - 1/2^2) eV -------(1)

for Be+3(z = 4)

Energy of photon emited with shortest wavelength or highest energy

E = 13.6*z^2*(1/nf^2 - 1/ni^2) eV

Emax = 13.6*4^2*(1/1^2 - 1/2^2) eV

h*c/lamda_Be = 13.6*4^2*(1/1^2 - 1/2^2) eV -------(1)

take equation(1)/equation(2)

lamda_Be/lamda_Li = 3^2/4^2

lamda_Be = lamda_Li*(3/4)^2

= 40.5*(3/4)^2

= 22.8 nm <<<<<<<------------------Answer