# What is potential difference for an electron accelerated to 0.5n m ?

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### 1 Answer

You need to evaluate the speed that is involved in equation of kinetic energy, using the following relation, such that:

`v = h/(lambda*m) `

Replacing `6.63*10^(-34)` for h, `9.11*10^(-31)` for m and the given wavelength of` 0.5nC = 0.5*10^(-9)C` yields:

`v = (6.63*10^(-34))/(0.5*10^(-9)*9.11*10^(-31))`

`v= 1.47*10^6 => v = 147*10^4 m/s`

You need to evaluate the kinetic energy such that:

`E_c = (m*v^2)/2 => E_c = (9.11*10^(-31)*147*10^8)/2`

`E_c = 98428.995*10^(-23) = 984.289 10^(-16) J`

Since `E_c = eV` yields that `eV = 984.289 10^(-16) J.`

**Hence, evaluating the requested potential difference yields **`eV = 984.289 10^(-16) J.`

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