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The value of Planck's constant is 6.63xx...

The value of Planck's constant is `6.63xx10^-34` Js. The speed of light is `3xx10^17nms^-1`. which value is closest to the wavelength in nanometer of a quantum of light with frequency of `6xx10^15s^-1`?

A

10

B

25

C

50

D

75

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • The value of Planck's constant is 6.63xx10^(-34)J*s . The speed of light is 3xx10^(17)nm*s^(-1) . Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6xx10^(15)s^(-1) -

    A
    50
    B
    75
    C
    10
    D
    25

  • The Young's modulus of a metal is 1.2xx10^11Pa and its interatomic force constant is 3.6xx10^-9Nm^-1 the average distance between atoms is

    A
    `3xx10^-16 m`
    B
    `3xx10^-18 m`
    C
    `3xx10^-20 m`
    D
    `3xx10^-20 cm`

  • Calculate the energy in joule correspending to light of wavelength 45nm. (Planck's constant h= 6.63xx10^-34 Js, speed of light c= 3xx10^8m/s ).

    A
    `6.67xx10^15`
    B
    `6.67xx10^11`
    C
    `4.42xx10^-15`
    D
    `4.42xx10^-18`

    • Similar Questions

    Explore conceptually related problems

    The velocity of a proton is 10^3 ms^-1 and mass of it is 1.67 xx10^(-27) kg , determine the Wavelength of that proton particle in nanometer unit (h = 6.63 xx 10^(-34) Js)

    The mass of an electron = 9.1 xx 10^(-31) kg and Planck's constant = 6.625 xx 10^(-34)J.s . What is the de Broglie wavelength (in Å ) of the electron if its velocity is 7.28 xx 10^(6) m.s^(-1) ?

    Wavelength of ultraviolet light is 3 xx 10^(-5) cm . What will be the energy of a photon of this light, in eV? (c = 3 xx 10^(10) m.s^(-1))

    Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . The number of photons emitted per second from a light source of power 40 W and wavelength 5893 Å

    Einstein's equation for photoelectric effect is E_("max") = hf - W_(0) , where h = Planck's constant = 6.625 xx 10^(-34) J.s, f = frequency of light incident on metal surface, W_(0) = work function of metal and E_("max") = maximum kinetic energy of the emitted photoelectrons. It is evident that if the frequency f is less than a minimum value f_(0) or if the wavelength lamda is greater than a maximum value lamda_(0) , the value of E_("max") would be negative, which is impossible. Thus for a particular metal surface f_(0) is the threshold frequency and lamda_(0) is the threshold wavelength for photoelectric emssion to take place. Again if the collector plate is ketp at a negative potential with respect to the emitter plate, the velocity of the photoelectrons would decrease. The minimum potential for which the velocity of the speediest electron becoes zero, is known as the stopping potential, the photoelectric effect stops for a potential lower than this. [velocity of light = 3xx 10^(8) m.s^(-1) , mass of an electron m = 9.1 xx 10^(-31) kg , charge of an electron, e = 1.6 xx 10^(-19)C Ultraviolet of wavelength 1800 Å is incident on the metal surface. The maximum velocity of the emitted photoelectron (in m.s^(-1) ) is

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