Calculate the energy associated with the 1st orbit of He+. What is the radius of the orbit?

Pls give the answer in steps..!

same ques??

  • -9

  energy of  1st  orbit of He+ ion=-13.6*4/1=5.44 ev/atom. 


radius=0.529 angstroms.is it right aiswarya.

  • 8

 he Rydberg formula can be extended for use with any hydrogen-like chemical elements.
1/ λ = R*Z^2 [ 1/n1^2 - 1/n2^2] 
where
λ is the wavelength of the light emitted in vacuum;
R is the Rydberg constant for this element; R 1.09737x 10^7 m-1
Z is the atomic number, for He, Z =2;
n1 and n2 are integers such that n1 < n2
The energy of a He+ 1s orbital is the opposite to the energy needed to ionize the electron that is
taking it from n = 1 (1/n1^2 =1) to n2 = ∞ (1/n2^2 = 0)
.: 1/ λ = R*Z^2 = 1.09737x 10^7*(2)^2 
λ = 2.278*10^-8 m
E = h*c/λ 
Planck constant h = 6.626x10^-34 J s
c = speed of light = 2.998 x 10^8 m s-1
E = (6.626x10^-34*2.998 x 10^8)/(2.278*10^-8) = 8.72*10^-18 J ion-1
Can convert this value to kJ mol-1:
(8.72*10^-18*6.022 x 10^23)/1*10^3 = 5251 kJ mol-1
Lit value: RP’s secret book: 5240.4 kJ mol-1 (difference is due to a small change in R going from H to He+)
So energy of the 1s e- in He+ = -5251 kJ mol-1

  • -15
Plzz give answer in steps
  • -10
Energy of orbit n, En = R(Z2 / n2 ) and radius, rn = ao n2 / Z
RH = 2.18 * 10-18 J; Z = 2; n = 1; ao = 52.9 pm
E​n = ​2.18 * 10-18 J (4/1) = 8.72 * ​10-18 J
rn = 52.9 pm * 1 / 2 = 26.45 pm
 
  • 58
The value of Rydberg's constant is -2.18 * 10^-18, not 2.18 * 10^-18. So answer is: -8.72 * 10^-18 J
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ggg
  • -10
ao what does this mean??????
  • -3
Jgcitcitciycoh
  • 0
Energy of orbit n, En = RH (Z2 / n2 ) and radius, rn = ao n2 / ZRH = 2.18 * 10-18 J; Z = 2; n = 1; ao = 52.9 pmE​n = ​2.18 * 10-18 J (4/1) = 8.72 * ​10-18 Jrn = 52.9 pm * 1 / 2 = 26.45 pm
  • 1
[-8.72X10-18J]
 
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Energy of orbit n, En = RH (22/n2) and radius,

in = a, n? / Z

RH = 2.18 * 1018 J; Z = 2; n = 1; ao = 52.9 pm

En = 2.18 * 1018J (4/1) = 8.72 * 10-18 J In = 52.9 pm *1/2 = 26.45 pm
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