The kinetic energy of an electron is 5eV. Calculate the de-Broglie wavelength associated with it (h=6.6×10^-34 Js, m_e =9.1×10^-31 kg)

• 5.47A^o

• 10.9A^o

• 2.7A^o

• None of these

122.

Radius of first Bohr orbit is r. What is the radius of 2nd Bohr orbit ?

• 8 r

• 2 r

• 4 r

• $2\sqrt{2} \mathrm{r}$

Copper has face-centred cubic ( FCC ) lattice with interatomic spacing equal to 2.54 A. The value of lattice constant for this lattice is

• 1.27 A^o

• 5.08 A^o

• 2.54 A^o

• 3.59 A^o

J.J. Thomson's cathode ray tube experiment demonstrated- that

• cathode rays are streams of negatively charged ions

• all the mass of an atom is essentially in the nucleus

• the e/m of electrons is much greater than the e/m of protons.

• the e/m ratio of the cathode ray particles changes when a different gas is placed in the discharged tube

Wavelength of light emitted from second orbit to first orbit in a hydrogen atom is

• 6563 A^o

• 4102 A^o

• 4861 A^o

• 1215 A^o

If λ₁ and λ₂ are the wavelengths of the first members of the Lyman and Paschen series respectively, then λ₁ : λ₂ is

• 1:3

• 1: 30

• 7: 50

• 7: 108

Minimum excitation potential of Bohr's first orbit in hydrogen atom is

• 3.6 V

• 10.2 V

• 13.6 V

• 3.4 V

Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atom on the ground state rarely excited by monochromatic radiation of photon 12.1 eV. The special line emitted by a hydrogen atom according to Bohr's theory will be

• one

• two

• three

• four

According to Bohr's model of hydrogen atom, relation between principal quantum number n and radius of stable orbit is

• r ∝ $\frac{1}{\mathrm{n}}$

• r ∝ n

• r ∝ $\frac{1}{{\mathrm{n}}^2}$

• r ∝ n²

The de-Broglie wavelength of electron falling on the target in an X-ray tube is),. The cut-off wavelength of the emitted X-ray is

• ${\mathrm{\lambda }}_0 = \frac{{\left( \mathrm{m} \mathrm{c} \mathrm{\lambda } \right)}^2}{\mathrm{h}}$

• ${\mathrm{\lambda }}_0 = \frac{{\mathrm{m}}^2 \mathrm{c\lambda }}{{\mathrm{h}}^2}$

• ${\mathrm{\lambda }}_0 = \frac{2 \mathrm{mc} {\mathrm{\lambda }}^2}{\mathrm{h}}$

• ${\mathrm{\lambda }}_0 = \frac{\mathrm{m} \mathrm{c} {\mathrm{\lambda }}^2}{{\mathrm{h}}^2}$