Advertisements
Advertisements
प्रश्न
The energy of hydrogen atom in an orbit is −1.51 eV. What are the kinetic and potential energies of the electron in this orbit?
Advertisements
उत्तर
Total energy = −1.51 eV
So, Potential energy = 2 × Total energy = −3.02 eV
Kinetic energy = −Total energy = 1.51 eV
APPEARS IN
संबंधित प्रश्न
Using Rutherford's model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
Answer the following question, which help you understand the difference between Thomson’s model and Rutherford’s model better.
Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
An electron in an atom revolves round the nucleus in an orbit of radius r with frequency v. Write the expression for the magnetic moment of the electron.
For 7.7 Mev alpha particles scattering from aluminium (Z = 13), the distance of closest approach in a bead on collision is ______.
Useful data
`1/(4 pi ∈_0) = 8.99 xx 10^9` newton m2C-2; c = 1.60 × 10-19 C; leV = 1.60 × 10-19j.
A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α- particles and 2 positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be:
The ratio of the frequencies of the long wave length its of Lyman Balmer series of hydrogen spectrum is
In 88 Ra226 nucleus, there are
O2 molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms ______.
Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.
The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge + q1, –q2 is modified to |F| = `(q_1q_2)/((4πε_0)) 1/r^2, r ≥ R_0 = (q_1q_2)/(4πε_0) 1/R_0^2 (R_0/r)^ε, r ≤ R_0` Calculate in such a case, the ground state energy of a H-atom, if ε = 0.1, R0 = 1Å.
