A single electron in an ion has ionisation energy equal to 217.6 eV. What is the total number of neutrons present in that ion is :

Question

A single electron in an ion has ionisation energy equal to 217.6 eV. What is the total number of neutrons present in that ion is : (A) 2 (B) 4 (C) 5 (D) 9

Tardigrade Admin · Accepted Answer

Ionisation energy of ion = E ∞- E 1=217.6 eV therefore E 1=-217.6 eV =-13.6 × ( Z 2/12) ∴ Z2=(217.6/13.6)=16 and Z=4 thus in berylium ( Z =4) ion no. of neutrons =9-4=5

Q. A single electron in an ion has ionisation energy equal to $217.6 e V$ . What is the total number of neutrons present in that ion is :

$2$

$4$

$5$

$9$

Ionisation energy of ion $= E_{\infty} - E_{1} = 217.6 e V$
therefore $E_{1} = - 217.6 e V = - 13.6 \times \frac{Z ^{2}}{1 ^{2}}$
$∴ Z^{2} = \frac{217.6}{13.6} = 16$ and $Z = 4$
thus in berylium $(Z = 4)$ ion no. of neutrons
$= 9 - 4 = 5$

Q. A single electron in an ion has ionisation energy equal to 217.6eV. What is the total number of neutrons present in that ion is :

2

4

5

9

Ionisation energy of ion =E∞​−E1​=217.6eV therefore E1​=−217.6eV=−13.6×12Z2​ ∴Z2=13.6217.6​=16 and Z=4 thus in berylium (Z=4) ion no. of neutrons =9−4=5

Q. A single electron in an ion has ionisation energy equal to $217.6 e V$ . What is the total number of neutrons present in that ion is :

$2$

$4$

$5$

$9$

Ionisation energy of ion $= E_{\infty} - E_{1} = 217.6 e V$
therefore $E_{1} = - 217.6 e V = - 13.6 \times \frac{Z ^{2}}{1 ^{2}}$
$∴ Z^{2} = \frac{217.6}{13.6} = 16$ and $Z = 4$
thus in berylium $(Z = 4)$ ion no. of neutrons
$= 9 - 4 = 5$