What is the energy of state in a triply ionised berrylium $(B e^{+ + +})$ whose radius is equal to that of ground state of hydrogen?

27.2 eV

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54.4 eV

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13.6 eV

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40.8 eV

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Solution

The correct option is B 54.4 eV
$r_{1} = r_{2}, \frac{Z_{1}}{Z_{2}} = {(\frac{n_{1}}{n_{2}})}^{2} A s r α (\frac{n^{2}}{z})$
$\frac{1}{4} = {(\frac{1}{n_{2}})}^{2}$
$n_{2} = 2$
$E = - 13.6 \frac{z^{2}}{n^{2}}$
$E = - 13.6 \times \frac{4^{2}}{2^{2}}$
$= - 13.6 \times 4$
$= 54.4 e V$

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What is the energy of state in a triply ionised berrylium (Be+++) whose radius is equal to that of ground state of hydrogen?

The correct option is B 54.4 eVr1=r2,Z1Z2=(n1n2)2 Asrα(n2z) 14=(1n2)2 n2=2 E=−13.6z2n2 E=−13.6×4222 =−13.6×4 =54.4eV

What is the energy of state in a triply ionised berrylium $(B e^{+ + +})$ whose radius is equal to that of ground state of hydrogen?

The correct option is B 54.4 eV
$r_{1} = r_{2}, \frac{Z_{1}}{Z_{2}} = {(\frac{n_{1}}{n_{2}})}^{2} A s r α (\frac{n^{2}}{z})$
$\frac{1}{4} = {(\frac{1}{n_{2}})}^{2}$
$n_{2} = 2$
$E = - 13.6 \frac{z^{2}}{n^{2}}$
$E = - 13.6 \times \frac{4^{2}}{2^{2}}$
$= - 13.6 \times 4$
$= 54.4 e V$