Question

For H-spectrum, electron transition takes place from
n-5 ton=2 then emitted wavelength of photon
is 434 nm. The wave length of photon in electron
transition from n = 4 to n = 2 will be :-o​

Answer 1

U probably know this formula,

1/lambda=Rz^2(1/n1^2-1/n2^2)

So putting 434 in the formula calculate R and then place the value of R in the next transition.

Answer 2

Given - n1 - 5, n2 - 2, lambda1 - 434nm, n3 - 4, n4 - 2

Find - lambda2

Answer - The formula to be used for the calculation of wavelength is given by Rydberg formula. The fromula is mentioned as follows -

1/lambda = Rz²(1/n1²-1/n2²)

In the formula, R represents Rydberg constant, n refers to quantum number.

For the calculation in given question is -

Lambda2/lambda1 = n3²n4²(n2²-n1²)/n1²n2²(n4²-n3²)

Lambda2/434 = 4²2²(2²-5²)/2²5²(2²-4²)

Lambda2/434 =16*4(4-25)/4*25(4-16)

Lambda2/434 = 64(-21)/100(-12)

Lambda2 = -1344/-1200

Lambda2 = 1.12*434

Lambda2 = 486.08 mm

Hence, wavelength for transition from n = 4 to n = 2 is 486 nm.