We know that,

 

 =2.303/t  log a ₀ /a

where lembda= disintegration constant

          t   = time

          a ₀= amount of radioactive substance at 0 time(in our case 100%)

          a   = substance after time t

2.303/1  log  100/90  =  2.303/2  log  100/a

cancelling 2.303 on both sides ,we get

2 log 100 - 2 log 90   = log 100 -log a

4 - 2 log 90 = 2 - log a

log a= 2 - 4  + 2 log 90

= -2  +2 (1.9542)

= -2  +3.9084

= 1.9084

a = antilog of 1.9084

= 80.98

hence, the pecentage of radioactive substance remaining at the end of two hours is 80.98%.