JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is:
Numerical answer
More from Atoms and Nuclei
- The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?2020 · MCQ
- Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation is :2019 · MCQ
- The ratio of mass densities of nuclei of 40Ca and 16O is close to :-2019 · MCQ
- Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :2019 · MCQ
- A He+ ion is in its first excited state. Its ionization energy is :-2019 · MCQ
- In Li+ +, electron in first Bohr orbit is excited to a level by a radiation of wavelength . When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are…2019 · MCQ
- Consider the nuclear fission Ne20 2He4 + C12 Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -2019 · MCQ
- A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 . The radius of the atom in the excited state, in terms of Bohr radius a0 will be : (hc = 12500 eV …2019 · MCQ