Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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.

Sagot :

Cricetus

Answer:

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

Explanation:

The given values are:

Potential difference,

Δv = 400 v

Radius,

r = 0.5580 cm

As we know,

⇒  [tex]W=e \Delta v[/tex]

and,

⇒  [tex]\frac{1}{2}mv^2=e \Delta v[/tex]

then,

⇒  [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]

On substituting the values, we get

⇒     [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]

⇒     [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]

⇒     [tex]=1.185\times 10^7 \ m/sec[/tex]

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.