The work function of caesium is 2.14 eV. The wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V will be:
454 nm
440 nm
333 nm
350 nm
Photons absorbed in matter are converted to heat. A source emitting n photon/sec of frequency ν is used to convert 1 kg of ice at 0°C to water at 0°C. Then, the time T taken for the conversion:
(a) decreases with increasing n, with ν fixed(b) decreases with n fixed, ν increasing(c) remains constant with n and ν changing such that nν=constant(d) increases when the product nν increases
(a, c, d)
(a, d)
(a, b, c)
The de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron. The speed of the electron is ve=c100. Then,
EeEp=10-4, pemec=10-2
EeEp=10-2, pemec=10-2
EeEp=10-2, pemec=10-4
EeEp=10-4, pemec=10-4
Two particles A1 and A2 of masses m1, m2 (m1>m2) have the same de-Broglie wavelength. Then,
(a) their momenta are the same(b) their energies are the same(c) energy of A1 is less than the energy of A2(d) energy of A1 is more than the energy of A2
(b, c)
(a, c)
(c, d)
(b, d)
Relativistic corrections become necessary when the expression for the kinetic energy 12mv2, becomes comparable with mc2, where m is the mass of the particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?
(a) λ=10 nm(b) λ=10-1 nm(c) λ=10-4 nm(d) λ=10-6 nm
(a, b)
An electron (mass m) with an initial velocity v→=v0i^ is in an electric field E→=E0j^. If λ0=hmv0, its de-Broglie wavelength at time t is given by:
λ0
λ01+e2E02t2m2v02
An electron (mass m) with an initial velocity v→=v0i^ (v0>0) is in an electric field E→=-E0i^ (E0=constant>0). Its de-Broglie wavelength at time t is given by:
λ0(1+eE0mtv0)
λ0(1+eE0tmv0)
λ0t
An electron is moving with an initial velocity v→=v0i^ and is in a magnetic field B→=B0j^. Then, its de-Broglie wavelength:
remains constant
increases with time
decreases with time
increases and decreases periodically
A proton, a neutron, an electron and an α-particle have the same energy. Then, their de-Broglie wavelengths compare as:
λp=λn>λe>λα
λα<λp=λn<λe
λe<λp=λn>λα
λe=λp=λn=λα
Monochromatic light of frequency 6.0×1014 Hz is produced by a laser. The power emitted is 2.0×10-3 W. How many photons per second, on average, are emitted by the source?
1. 6×10152. 5×10143. 6×10144. 5×1015
The wavelength of light in the visible region is about 390 nm for violet colour. The energy of photons in (eV) is:
Take h=6.63×10-34 Js, and 1 eV=1.6×10-19 J
4.03 eV
1.64 eV
2.26 eV
3.19 eV
What is the de-Broglie wavelength associated with an electron moving at a speed of 5.4×106 m/s?
4. 0.111 nm
2. 0.135 nm
3. 0.157 nm
1. 0.244 nm
The de Broglie wavelength associated with a ball of mass 150 g travelling at a speed of 30.0 m/s is:
1. 1.47×10-32 m2. 2.01×10-34 m3. 2.01×10-32 m4. 1.47×10-34 m
An electron, an α-particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?
Electron
α-particle
proton
All have the same de Broglie wavelength
A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813×10-4. The particle’s mass is:
1. 9.371×10-31 kg2. 9.001×10-31 kg3. 1.675×10-27 kg4. 2.705×10-27 kg
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts?
0.123 nm
0.232 nm
0.031 nm
0.312 nm
Graph shows variation of photoelectric current with intensity of light. It can be concluded that photoelectrons emitted per second are-
Independent of intensity of incident radiation
Inversely proportional to intensity of incident radiation
Directly proportional to intensity of incident radiation
Directly proportional to square of intensity of incident radiation
The graph shows the variation of photocurrent with collector plate potential for different intensities of incident radiation. Then the maximum intensity is:
I1
I2
I3
All intensities are same
The wavelength of light in the visible region is about 550 nm (average wavelength) for yellow-green colour. Three materials with work functions are given as Al (4.28 eV), Cu (4.65 eV) and Na (2.75 eV). From which of these photosensitive materials, can you build a photoelectric device that operates with visible light?
Al
Cu
Na
None of the above
The number of photons per second on an average emitted by a source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3×10-3 watt will be: h=6.6×10-34 Js
1016
1015
1018
1017
An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' is mass of photoelectron emitted from the surface has de-Broglie wavelength λd, then:
λ=2mchλd2
λ=2hmcλd2
λ=2mhcλd2
λd=2mchλ2
Please disable the adBlock and continue. Thank you.