If electron, hydrogen, helium and neon nuclei are all moving with the velocity of light, then the wavelength associated with these particles are in the order

Electron > hydrogen > helium > neon
Electron > helium > hydrogen> neon
Electron < hydrogen < helium < neon
Neon < hydrogen < helium < electron

Solution

λ = h/mv ; for the same velocity, λ varies inversely with the mass of the particle.

The position of both, an electron and a helium atom is known within 1.0 nm. Further the momentum of the electron is known within 5.0 × 10^–26kg ms^–1. The minimum uncertainty in the measurement of the momentum of the helium atom is

50 kg ms^–1
80 kg ms^–1
8.0 × 10^–26kg ms^–1
5.0 × 10^–26kg ms^–1

Solution

By Heisenberg uncertainty Principle Δx × Δp = ^h⁄_4π (which is constant)

As Δx for electron and helium atom is same thus momentum of electron and helium will also be same therefore the momentum of helium atom is equal to 5 × 10^–26 kg. m.s^–1.

Positron is :

electron with positive charge
a nucleus with one neutron and one proton
a nucleus with two protons
a helium nucleus

Solution

Positron is electron with positive charge, ₊₁e⁰

The configuration 1s², 2s² 2p⁵, 3s¹ shows :

excited state of O^-₂
excited state of neon atom
excited state of fluorine atom
ground state of fluorine atom

Solution

Atomic number of the given element = 10

Electronic configuration = 1s², 2s² p⁶

1s², 2s² 2p⁶ is electronic configuration of Ne.

1s², 2s² 2p⁵, 3s¹ is excited oxidation state.

Uncertainty in position of a n electron (mass = 9.1 × 10^–28g) moving with a velocity of 3 × 10⁴ cm/s accurate upto 0.001% will be (use h/4π) in uncertainty expression where h = 6.626 × 10^-27 erg-second).

1.93 cm
3.84 cm
5.76 cm
7.68 cm

Solution

In the photo-electron emission, the energy of the emitted electron is

greater than the incident photon
same as than of the incident photon
smaller than the incident photon
proportional to the intensity of incident photon

Solution

K.E. of emitted electron

= hv - hv₀ (i.e.smaller than hv).

If the energy of a photon is given as : = 3.03 × 10^–19J then, the wavelength (λ) of the photon is :

6.56 nm
65.6 nm
656 nm
0.656 nm

Solution

The ionisation potential of a hydrogen atom is –13.6 eV. What will be the energy of the atom corresponding to n = 2.

–3.4 eV
–6.8 eV
–1.7 eV
–2.7 eV

Solution

Energy of an electron E = ^-E₀⁄_n²

For energy level (n = 2)

E = ^-13.6⁄_(2)² = ^-13.6⁄₄ = -3.4 eV.

When atoms are bombarded with alpha particles, only, a few in million suffer deflection, others pass out undeflected. This is because

the force of repulsion on the moving alpha particle is small
the force of attraction between alpha particle and oppositely charged electrons is very small
there is only one nucleus and large number of electrons
the nucleus occupies much smaller volume compared to the volume of the atom

Solution

The nucleus occupies much smaller volume compared to the volume of the atom.

In a Bohr model of an atom, when an electron jumps from n = 3 to n = 1, how much energy will be emitted?

2.15 × 10^–11 ergs
2.389× 10^–12 ergs
0.239 × 10^–10 ergs
0.1936 × 10^–10 ergs

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