Which energy state of the triply ionized beryllium ( Be +++) has the same electron orbital radius as that of the ground state of hydrogen? Given Z for beryllium =4.

Question

Which energy state of the triply ionized beryllium ( Be +++) has the same electron orbital radius as that of the ground state of hydrogen? Given Z for beryllium =4. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As rn propto (n2/Z) we use ⇒ (n Be 2/Z Be )=(n H 2/Z Be ) ⇒ n Be 2=((1)2/1) × 4 ⇒ n Be =2

Q. Which energy state of the triply ionized beryllium $(B e^{+++})$ has the same electron orbital radius as that of the ground state of hydrogen? Given $Z$ for beryllium $= 4$ .

As $r_{n} \propto \frac{n ^{2}}{Z}$ we use
$\Rightarrow \frac{n _{B e}^{2}}{Z _{B e}} = \frac{n _{H}^{2}}{Z _{B e}}$
$\Rightarrow n_{B e}^{2} = \frac{( 1 ) ^{2}}{1} \times 4$
$\Rightarrow n_{B e} = 2$

Q. Which energy state of the triply ionized beryllium (Be+++) has the same electron orbital radius as that of the ground state of hydrogen? Given Z for beryllium =4.

As rn​∝Zn2​ we use ⇒ZBe​nBe2​​=ZBe​nH2​​ ⇒nBe2​=1(1)2​×4 ⇒nBe​=2

Q. Which energy state of the triply ionized beryllium $(B e^{+++})$ has the same electron orbital radius as that of the ground state of hydrogen? Given $Z$ for beryllium $= 4$ .

As $r_{n} \propto \frac{n ^{2}}{Z}$ we use
$\Rightarrow \frac{n _{B e}^{2}}{Z _{B e}} = \frac{n _{H}^{2}}{Z _{B e}}$
$\Rightarrow n_{B e}^{2} = \frac{( 1 ) ^{2}}{1} \times 4$
$\Rightarrow n_{B e} = 2$