Please answer this question?

eV = h [ν - ν0}

[ν2 - ν1] = C [1/λ2 - 1/ λ1] = [λ1- λ2] / λ1 λ2

= 3e8 [800-700] / 800*700* 10^-10 = = 5.357e+14

Difference in k.e = 4 -1.8 = = 2.2eV.

e = 1.602e-19 Coulomb

h= [diff.in k.e] / [ν2 - ν1]

h = 2.2* 1.602e-19/5.357e+14

h = 6.58e-34 Js.

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Formula meant for this problem is h(nu - nu not) = KE

nu-the frequency of radiation falling and nu not is the threshold frequency which is not given in the problem. So it is better to eliminate nu not.

Here nu is c/lambda where c is velocity of light ie 3 * 10^8 m/s and lambda the wavelength given.

wavelength in first case = 800 angstrom = 8*10 ^--8m

Frequency in first case = 3*10^8 / 8*10^--7 = (3/8)*10*14 Hz

Wavelength in the second case = 700 angstrom = 7*10^--8 m

Frequency in the second case = (3/7)*10^14 Hz

For font convenience let me replace nu by f.

So in the first case, h(f1 - fo) = KE

ie h((3/8)*10*14 -- f 0) = 1.8 eV ----------->(1)

Same way in the second case, h((3/7)*10^14 -- fo) = 4 eV---------->(2)

As we subtract then the threshold frequency fo will get cancelled automatically, which will be needed.

So subtracting (1) from (2), we get, h(3/7--3/8)*10^14 = 2.2 eV

Here we have to convert eV into J by multiplying with 1.602*10^--19

h(3/56)*10^14 = 2.2*1.602*10^--19

So with all substitutions, and solving we get, h = 6.57888*10^ --34 Js

Note: The actual value of h is 6.626 * 10^ --34 Js.

So the chapter is about Plank's constant.I heard it somewhere.I haven't yet studied Planck's constant up to now, I think I'm going to study it.

Check out the source section for the sites I suggest you to look at.