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 162.2 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength?

Answer 1

we have

shortest wavelength corresponds to the fall of electron from the n^th orbital to the free state

so we can write an as follows

1 / = Z² x 1.097373 x 10⁷ ( 1 / n² - 1 / )

1 / ( 162.2 x 10^-9 ) = 9 x 1.097373 x 10⁷ ( 1 / n² - 0 )

1/n² = 0.06242

n = 4

for beryllium's shortest wavelength is

1 / = Z² x 1.097373 x 10⁷ ( 1 / n² - 1/ )

1 / =16 x 1.097373 x 10⁷ ( 1 / 4² - 0)

1 / = 16 x 1.097373 x 10⁷ ( 1 / 16 )

1 / = 1.097373 x 10⁷

= 9.1429 x 10^-8 m

= 91.429 x 10^-9 m

so the shortest wavelength is = 91.429 nm