The De Broglie Explanation
Trending Questions
- λ2e∝λp
- λp∝λe
- λp∝√λe
- λp∝√λe
Calculate the wavelength of the first line in the Lyman series of the hydrogen spectrum () how to do this?
- v3c
- 2cv
- v2c
- 3cv
- 25
- 75
- 60
- 50
(Given: h=6.63×10−34 Js, me=9.11×10−31kg, e=1.6×10−19coulomb
- 725 pm
- 500 pm
- 322 pm
- 112 pm
If and are the wavelengths of the third member of Lyman and the first member of the Paschen series respectively, then the value of is
- E−1
- E0
- E1/2
- E−2
- 34
- 14
- 18
- 38
- electron
- proton
- CO2 molecule
- NH3 molecule
- 802 nm
- 823 nm
- 1882 nm
- 1648 nm
(Speed of light, c=3×108 ms−1Planck's constant, h=6.63×10−34 JsMass of electron, me=9.1×10−31 kg)
- 1.1×106 ms−1
- 1.7×106 ms−1
- 1.8×106 ms−1
- 1.45×106 ms−1
What is the quantum effect?
- E2
- E−1/2
- E−1
- E1/2
What is the speed of electron in the second Bohr's orbit of Hydrogen atom? (Take v0=2.2×106 m/s).
According to De Broglie 2πr=nλ where n is the quantum number of an orbit, in visual representation of an electron as a wave, n stands for:
No. of loops , where one loop is one wavelength
No. of electrons
No. of half wavelengths
Can not comment
For the Balmer series in the spectrum of H-atom, The correct statements among (A) to (D) are:
I) The integer
II) The ionization energy of hydrogen can be calculated from the wave number of these lines.
III) The lines of longest wavelength correspond to .
IV) As wavelength decreases, the lines of the series converge.
II, III, IV
I, II, IV
I, III, IV
I, II, III
- 3.33∘A
- 106∘A
- 1.67∘A
- 0.53∘A
- 5.47 ˚A
- 10.9 ˚A
- 2.7 ˚A
- None of these
- 2S
- S
- 3S
- S2
- Total energy in the nth orbit is proportional to n6
- Total energy in the nth orbit is proportional to m−3 (m is mass of electron)
- Total energy in the nth orbit is proportional to n−2
- Total energy in the nth orbit is proportional to m3 (m is mass of electron)
The circumference of the second orbit of an atom or ion having a single electron is The de-Broglie wavelength of electrons revolving in this orbit should be
- radius of the orbit
- perimeter of the orbit
- diameter of the orbit
- half of the perimeter of the orbit
(h=6.6×10−34J−s, me=9.1×10−31kg)
- 5.47˚A
- 10.9˚A
- 2.7˚A
- None of these
- Energy in the nth orbit is propotional to n6
- Energy is proportional to m−3 (m= mass of electron)
- Energy in the nth orbit is proportional to n−2
- Energy is proportional to m3 (m= mass of electron)
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of , the minimum electron energy required is close to :
- 3.315×10−28 kg-m/s
- 1.66×10−28 kg-m/s
- 4.97×10−28 kg-m/s
- 9.9×10−28 kg-m/s