Question

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10−2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . The potential difference between the anode and cathode is 400 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.

Answer 1

Answer:

The final speed will be "$1.185\times 10^7 \ m/sec$".

Explanation:

The given values are:

Potential difference,

Δv = 400 v

Radius,

r = 0.5580 cm

As we know,

⇒  $W=e \Delta v$

and,

⇒  $\frac{1}{2}mv^2=e \Delta v$

then,

⇒  $v=\sqrt{\frac{2e \Delta v}{m} }$

On substituting the values, we get

⇒     $=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }$

⇒     $=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}$

⇒     $=1.185\times 10^7 \ m/sec$