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) from Chemistry Structure of Atom Class 12 Nagaland Board
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Class 10 Class 12

The angular momentum of electrons in d orbital is equal to

  • square root of 6 space h
  • square root of 2 straight h
  • 2 square root of 3 straight h
  • 0 h


A.

square root of 6 space h

Angular momentum of electrons in d- orbital is 

square root of straight l space left parenthesis straight l plus 1 right parenthesis end root space fraction numerator h over denominator 2 pi end fraction semicolon space f o r space d space minus o r b i t a l space space l space equals 2
equals space square root of 2 space left parenthesis 2 plus 1 right parenthesis end root space equals space h space equals space square root of 6 h end root space space space space open parentheses therefore space h equals space fraction numerator h over denominator 2 pi end fraction close parentheses space

Angular momentum of electrons in d- orbital is 

square root of straight l space left parenthesis straight l plus 1 right parenthesis end root space fraction numerator h over denominator 2 pi end fraction semicolon space f o r space d space minus o r b i t a l space space l space equals 2
equals space square root of 2 space left parenthesis 2 plus 1 right parenthesis end root space equals space h space equals space square root of 6 h end root space space space space open parentheses therefore space h equals space fraction numerator h over denominator 2 pi end fraction close parentheses space

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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 10-18


D.

4.42 x 10-18

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

straight E space equals hc over straight lambda
Given comma
straight lambda space equals space 45 space mm space equals space 45 space straight x space 10 to the power of negative 9 end exponent space straight m space
left square bracket therefore space 1 space nm space equals space 10 to the power of negative 9 end exponent space straight m space right square bracket

Hence comma space straight E equals fraction numerator 6.63 space straight x space 10 to the power of negative 34 end exponent space js space straight x space 3 space straight x space 10 to the power of 8 space ms to the power of negative 1 end exponent over denominator 45 space straight x space 10 to the power of negative 9 end exponent space straight m end fraction
space equals space 4.42 space straight x space 10 to the power of negative 18 end exponent space straight j

Hence comma space the space energy space corresponds space to space light space of space wavelenght space 45 space nm space is space 4.42 space straight x space 10 to the power of negative 18 space end exponent space straight j

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

straight E space equals hc over straight lambda
Given comma
straight lambda space equals space 45 space mm space equals space 45 space straight x space 10 to the power of negative 9 end exponent space straight m space
left square bracket therefore space 1 space nm space equals space 10 to the power of negative 9 end exponent space straight m space right square bracket

Hence comma space straight E equals fraction numerator 6.63 space straight x space 10 to the power of negative 34 end exponent space js space straight x space 3 space straight x space 10 to the power of 8 space ms to the power of negative 1 end exponent over denominator 45 space straight x space 10 to the power of negative 9 end exponent space straight m end fraction
space equals space 4.42 space straight x space 10 to the power of negative 18 end exponent space straight j

Hence comma space the space energy space corresponds space to space light space of space wavelenght space 45 space nm space is space 4.42 space straight x space 10 to the power of negative 18 space end exponent space straight j

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What is the maximum number of orbitals that can be identified with the following quantum numbers?
n=3, l =1, m1 = 0

  • 1

  • 2

  • 3

  • 4


A.

1

The value of n=3 and l =1 suggest that it is a 3p orbital while the value of m1 = 0 [magnetic quantum number] shows that the given 3p orbital is 3pz in nature.
Hence, the maximum number of orbitals identified by the given quantum number is only 1, i.e. 3pz.

The value of n=3 and l =1 suggest that it is a 3p orbital while the value of m1 = 0 [magnetic quantum number] shows that the given 3p orbital is 3pz in nature.
Hence, the maximum number of orbitals identified by the given quantum number is only 1, i.e. 3pz.

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