Number of undecayed atom after time t2 ;

Given: Half-life of a radioactive substance = 20 minutes

To find: Time interval (t2 – t1) between the time t2 when ⅔ of it had decayed and time t1 when ⅓ of it had decayed.

Let's assume the initial amount of the radioactive substance to be 100 units.

After one half-life, the amount of substance remaining = 50 units.

After two half-lives, the amount of substance remaining = 25 units.

After three half-lives, the amount of substance remaining = 12.5 units.

Now, we need to find the time interval (t2 – t1) between the time t2 when ⅔ of it had decayed and time t1 when ⅓ of it had decayed.

⅓ of the initial amount = (1/3) × 100 = 33.33 units

⅔ of the initial amount = (2/3) × 100 = 66.67 units

When ⅓ of the initial amount decays, the remaining amount = 100 – 33.33 = 66.67 units.

When ⅔ of the initial amount decays, the remaining amount = 33.33 units.

From the above, we can see that the amount of substance remaining at time t1 is 66.67 units and at time t2 is 33.33 units.

Now, we need to find the time interval (t2 – t1).

From the formula for half-life, we know that:

Amount remaining after n half-lives = (1/2)^n × Initial amount

Amount remaining after t minutes = (1/2)^(t/20) × Initial amount

Let's assume that t1 is the time at which ⅓ of the initial amount has decayed.

Amount remaining at time t1 = (1/3) × Initial amount = (1/3) × 100 = 33.33 units

(1/2)^(t1/20) × 100 = 33.33

t1/20 = log(33.33/100) / log(1/2)

t1/20 = 1.585

t1 = 31.7 minutes (approx)

Similarly, let's assume that t2 is the time at which ⅔ of the initial amount has decayed.

Amount remaining at time t2 = (2/3) × Initial amount = (2/3) × 100 = 66.67 units

(1/2)^(t2/20) × 100 = 66.67

t2/20 = log(66.67/100) / log(1/2)

t2/20 = 0.585

t2 = 11.7 minutes (approx)

Therefore, the time interval (t2 – t1) = t2 – t1 = 11.7 – 31.7 = -20 minutes (approx)

However, since time cannot be negative, we take the absolute value of the result, which gives:

Time interval (t2 – t1) = |-20| = 20 minutes.

Hence, the correct option is (B) 20 min.