Advertisements
Advertisements
Question
The wavelength of the second line of the Balmer series in the hydrogen spectrum is 4861 Å. Calculate the wavelength of the first line of the same series.
Advertisements
Solution
For the first line in Balmer series:
`1/lambda = R(1/2^2 - 1/3^2) = (5R)/36` ..........(i)
For second Balmer line:
`1/4861 = R(1/2^2 - 1/4^2) = (3R)/16` ............(ii)
Dividing equation (ii) by (i),
`lambda/4861 = (3R)/16 xx 36/(5R)`
`lambda = 4861 xx 27/20 = 6562` Å
APPEARS IN
RELATED QUESTIONS
State Bohr’s third postulate for hydrogen (H2) atom. Derive Bohr’s formula for the wave number. Obtain expressions for longest and shortest wavelength of spectral lines in ultraviolet region for hydrogen atom
Find the frequency of revolution of an electron in Bohr’s 2nd orbit; if the radius and speed of electron in that orbit is 2.14 × 10-10 m and 1.09 × 106 m/s respectively. [π= 3.142]
On the basis of Bohr's theory, derive an expression for the radius of the nth orbit of an electron of the hydrogen atom.
Using Bohr's postulates, derive the expression for the orbital period of the electron moving in the nth orbit of hydrogen atom ?
The value of angular momentum for He+ ion in the first Bohr orbit is ______.
According to Bohr atom model, in which of the following transitions will the frequency be maximum?
Oxygen is 16 times heavier than hydrogen. Equal volumes of hydrogen and oxygen are mixed. The ratio of speed of sound in the mixture to that in hydrogen is ______.
On the basis of Bohr's theory, derive an expression for the radius of the nth orbit of an electron of hydrogen atom.
Calculate the energy associated with third orbit of He+.
The radius of hydrogen atom in the ground state is 0.53 Å. The radius of Li2+ ion (atomic number = 3) in a similar state is ______.
