A hydrogen like species in fourth orbit has radius 1.5 times that of Bohr’s orbit. In neutral state, its valence electron is in

Question

A hydrogen like species in fourth orbit has radius 1.5 times that of Bohr’s orbit. In neutral state, its valence electron is in (A) 3d (B) 3s (C) 2p (D) 1s

Tardigrade Admin · Accepted Answer

rn=(n2 a0/Z) Z=(n2 a0/rn)=(16 a0/1.5 a0)=10.67=11 Thus, Z=11 1 s 2 2 s 2 2 p 6 3 s 1 valence electron in 3 s

Q. A hydrogen like species in fourth orbit has radius 1.5 times that of Bohr’s orbit. In neutral state, its valence electron is in

3d

3s

2p

1s

$r_{n} = \frac{n ^{2} a _{0}}{Z}$

$Z = \frac{n ^{2} a _{0}}{r _{n}} = \frac{16 a _{0}}{1.5 a _{0}} = 10.67 = 11$

Thus, $Z = 11$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ valence electron in $3 s$

Q. A hydrogen like species in fourth orbit has radius 1.5 times that of Bohr’s orbit. In neutral state, its valence electron is in

3d

3s

2p

1s

rn​=Zn2a0​​ Z=rn​n2a0​​=1.5a0​16a0​​=10.67=11 Thus, Z=11 1s22s22p63s1 valence electron in 3s

$r_{n} = \frac{n ^{2} a _{0}}{Z}$

$Z = \frac{n ^{2} a _{0}}{r _{n}} = \frac{16 a _{0}}{1.5 a _{0}} = 10.67 = 11$

Thus, $Z = 11$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ valence electron in $3 s$