Grade 11Physical Chemistry
The velocity of electron in third ecicted state of be+3 will be
1 Answer
Priyanshu kumar
We know that, velocity of an electron in any Bohr orbit is V=Vo ×Z/n , where z= atomic no. of element and n= no. of Bohr orbit and Vo =2.16×10^6 ms^-1So in this case , z=4 (Beryllium) and n=4(3rd excited state).So, putting all values, V=2.16×10^6 × 4/4 = 2.16×10^6 ms^-1
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