Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the `2^(nd)` Balmer line (n = 4 to n = 2) will be :
Taking the wavelength of first Balmer line in the hydrogen spectrum (n=3" to "n=2) as 660 nm, then the wavelength of 2^("nd") Balmer line in the same spectrum (n=4" to "n=2) will be
The wavelength of the second line of balmer series in the hydrogen spectrum is 4861A^(@) .the wavelength of the first line is ?
If the wavelength of the first member of Balmer series of hydrogen spectrum is 6564A^(@) , the wavelength of second member of Balmer series will be:
Calculate the wavelength of the first line in the balmer series of hydrogen spectrum
If the wavelength of the first line of the Balmer series of hydrogen atom is 656.1 nm the wavelngth of the second line of this series would be
If the wavelength of the first member of Balmer series in hydrogen spectrum is 6563 Å, calculate the wavelength of the first member of Lymen series in the same spectrum.
