Solved 2024 Question Paper ICSE Class 10 Physics — Question 27

As only neutrons are involved in the reaction which number of protons and total nucleon number (i.e., number of protons + neutrons) remain constant.

Then,

Initially,

No. of protons (n_p) = Atomic number of $_{92}\text{U}^{235}$ = 92

No. of nucleons (n) = Mass number of $_{92}\text{U}^{235}$ + mass number of $_0\text n^1$ = 235 + 1 = 236

Now,

Finally,

Let atomic number of Kr is Z and mass number of Ba is A.

No. of protons (n_p) = Atomic number of $_{56}\text{Ba}^{\text A}$ + Atomic number of $_{\text Z}\text{Kr}^{92}$ = 56 + Z

and

No. of nucleons (n) = mass number of $_{56}\text{Ba}^{\text A}$ + mass number of $_{\text Z}\text{Kr}^{92}$ + 3 x mass number of $_0\text n^1$ = A + 92 + 3 = A + 95

As number of protons are constant :

⇒ 92 = 56 + Z

⇒ Z = 92 - 56 = 36

Also, number of nucleons are constant :

⇒ 236 = A + 95

⇒ A = 236 - 95 = 141

So reaction will be :

$_{92}\text{U}^{235} + _0\text{n}^1 \longrightarrow _{56}\text {Ba}^{141} + \space _{36}\text {Kr}^{92} + 3 \space _0\text n^1$