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Hydrogen Atom and Hydrogen Molecule Th...

Hydrogen Atom and Hydrogen Molecule
The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by johann Jakob Balmer, a Swiss teacher. Balmer's empirical empirical formula is
`1/lambda=R_(H)(1/2^(2)-1/n^(2))," " N=3, 4, 5`
`R_(H)=(Me e^(4))/(8 epsilon_(0)^(2)h^(3) c)=109. 678 cm^(-1)`
Here,
`R_(H)` is the Rydberg Constant, `m_(e)` is the mass of electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom//ion. the wavelength of first line of balmer region.

A

4340 Å

B

4101 Å

C

6569 Å

D

4861 Å

Text Solution

Verified by Experts

The correct Answer is:
C
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Hydrogen Atom and Hydrogen Molecule The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by johann Jakob Balmer, a Swiss teacher. Balmer's empirical empirical formula is 1/lambda=R_(H)(1/2^(2)-1/n^(2))," " N=3, 4, 5 R_(H)=(Me e^(4))/(8 epsilon_(0)^(2)h^(3) c)=109. 678 cm^(-1) Here, R_(H) is the Rydberg Constant, m_(e) is the mass of electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom//ion. Calculate the iongest wavelength in A (1 Å=10^(-10) m) in the 'Balmer series of singly ionized helium (He^(+)) Ignore nuclear motion in your calculation.

Hydrogen Atom and Hydrogen Molecule The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by johann Jakob Balmer, a Swiss teacher. Balmer's empirical empirical formula is 1/lambda=R_(H)(1/2^(2)-1/n^(2))," " N=3, 4, 5 R_(H)=(Me e^(4))/(8 epsilon_(0)^(2)h^(3) c)=109. 678 cm^(-1) Here, R_(H) is the Rydberg Constant, m_(e) is the mass of electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom//ion. From the given data, calculate the energy change for the process H^(-) rarr H+e^(-)

Hydrogen Atom and Hydrogen Molecule The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by johann Jakob Balmer, a Swiss teacher. Balmer's empirical empirical formula is 1/lambda=R_(H)(1/2^(2)-1/n^(2))," " N=3, 4, 5 R_(H)=(Me e^(4))/(8 epsilon_(0)^(2)h^(3) c)=109. 678 cm^(-1) Here, R_(H) is the Rydberg Constant, m_(e) is the mass of electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom//ion. A formula analogous to Balmer's formula applies to the series of spectral lines which arise from transition from higher energy levels to the lowest energy level of hydrogen atom. Write this formula and use it to determine the ground state energy of a hydrogen atom in eV. A 'muonic hydrogen atom' is like a hydrogen atom in which the electron is replaced by a heavier particle, the muon. The mass of a muon is about 207 times the mass of an electron, while its charge is the same as that of an electron. A muon has a very short lifetime, but we ignore its ubstable nature here.

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