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Calculate the wavelength of electron in ...

Calculate the wavelength of electron in an second orbit of `Be^(3+)` ion having radius equal to the Bohr's radius `(s_(0))`.

A

`1.34 A^(@)`

B

`1.53 A^(@)`

C

`1.66 A^(@)`

D

`1.46 A^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
In the expression, `r_(a) = (n^(2))/(Z^(2))a_(0)`, substituting Z=4 and `r=a_(0)`, we get `a_(0) =(n^(2))/(4^(2)) a_(0) rArr n=2`
Now, the ionization enthalpy is `E_(a) =-(Z^(2))/(n^(2))xx13.6 eV =(-4^(2))/(2^(2)) xx 13.6=-54.4 eV.`
So, the kinetic energy `=-E_(a) =54.4 eV`. Thus, the wavelength can be calculated as
`lamda =(h)/(sqrt(2mE)) =(6.626xx10^(-34))/(sqrt(2xx9.11xx10^(-31)xx54.4xx1.6xx10^(-19)))=1.66xx10^(-10)=1.66 Å`
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