The value of Planck's constant is 6.63 x 10-34 Js. The speed of light is 3 x1017 nms-1 . Which value is closet to the wavelength in nanometer of a quantum of light with frequency of 6 x 1015 s-1 ?

  • 10

  • 25

  • 50

  • 50


C.

50

Given, Planck's constant,
h= 6.63 x10-34
speed of light, c= 3 x1017 nms-1 
Frequency of quanta
v=6 x1015 s-1
Wavelength, λ =?

We know that,

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Maximum number of electrons in a subshell with l =3 and n=4 is

  • 14

  • 16

  • 10

  • 10


A.

14

n represents the main energy level and l represents the subshell.
If n=4 and l = 3, the subshell is 4f.
If f subshell, there are 7 orbitals and each orbital can accommodate a maximum number of electrons, so,  maximum number of electrons in 4f subshell = 7 x 2 = 14

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Calculate the energy in joule corresponding to light of wavelength 45mm: (Planck's constant h=6.63 x10-34)Js; speed of light c= 3 x 108 ms-1)

  • 6.67 x 1015

  • 6.67 x 1011

  • 4.42 x 10-15

  • 4.42 x 10-15


D.

4.42 x 10-15

The wavelength of light is related to its energy by the equation,

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Based on equation
E=-2.178 x 10-18open parentheses straight Z squared over straight n squared close parentheses certain conclusions are written. Which of them is not correct?

  • The negative sign in the equation simply means that the energy of an electron bound to the nucleus is lower than if would be if the electrons were at the infinite distance from the nucleus.

  • Larger the value of n, the larger is the orbit radius

  • Equation can be used to calculate the change in energy when the electron changes orbit

  • Equation can be used to calculate the change in energy when the electron changes orbit


D.

Equation can be used to calculate the change in energy when the electron changes orbit

If n=1,

E1 = - 2.178 x 10-18 Z2 J

If n=6

From the above calculation, it is obvious that electron has a more negative energy than it does for n=6. it means that electron is more strongly bound in the smallest allowed orbit.

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What is the maximum number of electrons that can be associated with the following set of quantum number?
n=3, l =1 and m=-1.

  • 10

  • 6

  • 4

  • 4


D.

4

The orbital of the electron having =3, l =1 and m= -1 is 3pz (as nlm) and an orbital can have a maximum of two electrons with opposite spins.
therefore, 3pz orbital contains only two electrons or only 2 electrons are associated with n=3, l=1, m=-1.

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