IF `rho` is the resistivity at temperature T, then the temperature coefficient of resistivity is defined as `alpha=(1drho)/(rhodT)`, which is a constant physical quantity for a given metal, show that `rho=rho_0e^(a(T-T_0)),` where `rho_0` = resistivity at temperature `T_0`.

Explanation

It is given that temperature coefficient of resistivity,
`alpha=1/rho(drho)/(dT)or,(drho)/rho=a,dT`
Integrating both sides, we get
`int(drho)/rho=inta.dT`
or, `log_erho=aT+k`……….(1)
where k= integration constant.
When `T=T_0,rho=rho_0`
`therefore` From equation (1), `k=log_erho_0-aT_0`
Putting k in equation (1), `log_erho=aT+log_erho_0-aT_0`
or, `log_e""rho/rho_0-a(T-T_0)or,rho/rho_0=e^(a(T-T_0))`
`thereforerho=p_0e^(a(T-T_0))`