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In the Bohr model of hydrogen atom. What is the de-Broglie wavelength lambda for the electron when it is in the (i) n = 1 level (ll) n = 4 level. In each case, compare the de-Broglie wave length to the circumference of the orbit. Data: n=1, n=4,lambda=? |
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Answer» Solution :By Bohr.s first POSTULATE, angular momentum of the electron `=(nh)/(2pi)` `mvr=(nh)/(2pi)` `:. (2pir)/(n)=(h)/(mv)` ...(1) But, (de brogile wave length), `lambda=(h)/(mv)` ...(2) From the equation (1) and (2) `lambda=(2pir)/(n)` ...(3) (i) `n=1, R=r_(1)=0.53 Å` `:. lambda_(1)=2pir_(1)=2pi(0.53)Å=3.328Å` From equation(3), `(lambda)/(2pir)=(1)/(n),` for n=1, `(lambda)/(2pir)=1`, i.e., `lambda-2pir` `:.` In first ORBIT, de-brogile wave length of the electron is equal to the circumference of the orbit. (ii) n = 4, `r_(4)=4^(2)r_(1)` from equation (3), `lambda_(4)=(2pir_(4))/(4)=(2pi(4^(2)r_(1)))/(4)` i.e., `lambda_(4)=2pixx4xx(0.53)Å` `lambda_(4)=13.313Å` and from `lambda_(4)=(2pir_(4))/(4)`, the de-Brogile wavelength of the electron in the fourth orbit is equal to one fourth of the PERIMETER. |
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