If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor 
(a) 12              (b) 2
(c) 12                 (d) 2

Concept Questions :-

De-broglie wavelength
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The energy that should be added to an electron to reduce its de Broglie wavelength from one nm to 0.5 nm is

(a) Four times the initial energy

(b) Equal to the initial energy

(c) Twice the initial energy

(d) Thrice the initial energy

Concept Questions :-

De-broglie wavelength
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The wavelength of the matter wave is independent of

(a) Mass                 (b) Velocity
(c) Momentum        (d) Charge

Concept Questions :-

De-broglie wavelength
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The energy of a photon of wavelength λ is given by
(a) hλ                       (b) chλ
(c) λ/hc                    (d)  hc/λ

Concept Questions :-

De-broglie wavelength
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The rest mass of the photon is

(a) 0
(b)
(c) Between 0 and
(d) Equal to that of an electron

Concept Questions :-

Particle nature of light
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The momentum of the photon of wavelength 5000Å will be

(a) 1.3×10-27 kg-m/sec         (b) 1.3×10-28 kg-m/sec
(c) 4×1029 kg-m/sec              (d) 4×10-18 kg-m/sec

Concept Questions :-

De-broglie wavelength
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The momentum of a photon of energy hv will be

(a) hv      (b) hv/c
(c) hvc     (d) h/v

Concept Questions :-

De-broglie wavelength
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If the momentum of a photon is p, then its frequency is

(a) phc           (b) pch

(c) mhc           (d) mch

where m is the rest mass of the photon

Concept Questions :-

De-broglie wavelength
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An AIR station is broadcasting the waves of wavelength 300 metres. If the radiating power of the transmitter is 10 kW, then the number of photons radiated per second is

(a) 1.5×1029              (b) 1.5×1031
(c) 1.5×1033              (d) 1.5×1035

Concept Questions :-

Electron emission
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The energy of a photon is E = hv and the momentum of photon p=hλ, then the velocity of photon will be

(a) E/p                      (b) Ep

(c)  EP2                  (d) 3×108 m/s

Concept Questions :-

De-broglie wavelength
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