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

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

Text Solution

Verified by Experts

The correct Answer is:
A
Wavelength of light, `lamda=(c)/(lamda)=(3xx10^(17))/(6xx10^(15))=50` nm
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Knowledge Check

  • 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

  • H, He^(+), Li^(2+) are examples of atoms or ions with one electron each . The energy of such atoms when in the n-th energy state (according to Bohr,s theory , n=1,2,3…. =principal quantum number ) is E_n =(-13.6 Z^2)/(n^2) eV (1 eV =1.6xx10^(-19)J) . For the ground state ,n=1 . in order to raise the atom from the ground state to n=f , the suitable incident light should have a wavelength given by lambda=(hc)/(E_f-E_1) . But the atom cannot stay permanently in the f-energy state, ultimately , it comes to the ground state by radiating the extra energy , E_f-E_1 as electromagnetic radiation . The electron of the atom comes from n=f to n=1 in one or more steps using the permitted energy levels . As a result there is a possibility of emission of radiation with more than one wavelength from the atom. Planck's constant =6.63 xx10^(-34)J*s and velocity of light c=3xx10^(8)m*s^(-1) . (i)What is the wavelength of the light incident on the atom to raise it to the fourth quantum level from ground state ?

    A
    952 Å
    B
    975 Å
    C
    1027 Å
    D
    1219 Å

  • Refractive index of a medium in which velocity of light is 2 xx 10^(8) m * s^(-1) , is

    A
    1.4
    B
    2.3
    C
    1.5
    D
    `1.0`

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