If the velocity of an electron moving in first Bohr's orbit of hydrogen has to be reduced to 7.27 × 105 m / s , it should be shifted by

Question

If the velocity of an electron moving in first Bohr's orbit of hydrogen has to be reduced to 7.27 × 105 m / s , it should be shifted by (A) 4.23 mathringA away from nucleus (B) 4.23 mathringA toward nucleus (C) 4.76 mathringA away from nucleus (D) 4.76 mathringA toward nucleus

Tardigrade Admin · Accepted Answer

7.27 × 105 m / s is the velocity of electron in the third Bohr's orbit of hydrogen atom. It means the electron has to be sent to the third shell. rn=0.529 n2 mathringA for hydrogen. Therefore, the distance through which an electron has to be shifted away from the nucleus =r3-r1. =[0.529 ×(3)2-0.529 ×(1)2] mathringA =4.23 mathringA

Q. If the velocity of an electron moving in first Bohr's orbit of hydrogen has to be reduced to $7.27 \times 1 0^{5} m / s,$ it should be shifted by

$4.23 \overset{˚}{A}$ away from nucleus

$4.23 \overset{˚}{A}$ toward nucleus

$4.76 \overset{˚}{A}$ away from nucleus

$4.76 \overset{˚}{A}$ toward nucleus

Q. If the velocity of an electron moving in first Bohr's orbit of hydrogen has to be reduced to 7.27×105m/s, it should be shifted by

4.23A˚ away from nucleus

4.23A˚ toward nucleus

4.76A˚ away from nucleus

4.76A˚ toward nucleus

Q. If the velocity of an electron moving in first Bohr's orbit of hydrogen has to be reduced to $7.27 \times 1 0^{5} m / s,$ it should be shifted by

$4.23 \overset{˚}{A}$ away from nucleus

$4.23 \overset{˚}{A}$ toward nucleus

$4.76 \overset{˚}{A}$ away from nucleus

$4.76 \overset{˚}{A}$ toward nucleus