The ionisation potential of mercury is 10.39 V. How far an electron mu

1.	The ionisation potential of mercury is 10.39 V. How far an electron must travel in an electric field of `1.5 xx 10^(6) V//m` to gain sufficient energy to ionise mercury ?A. `(10.39)/(1.6xx10^(-19))` mB. `(10.39)/(2xx1.6xx10^(-19))` mC. `10.39 xx 1.6 xx 10^(-19)` mD. `(10.39)/(1.5xx10^(6))` m
Answer» Correct Answer - D E=dV/dx ......................................eq 1 here E=1.5 x \(10^6\) V/m dV=10.39 V we have to find dx so substituing in eq 1,we get 1.5 x\(10^6\) = 10.39/dx dx=10.39/1.5 x \(10^6\) m so the answer is option d

The ionisation potential of mercury is 10.39 V. How far an electron must travel in an electric field of `1.5 xx 10^(6) V//m` to gain sufficient energy to ionise mercury ?A. `(10.39)/(1.6xx10^(-19))` mB. `(10.39)/(2xx1.6xx10^(-19))` mC. `10.39 xx 1.6 xx 10^(-19)` mD. `(10.39)/(1.5xx10^(6))` m

Answer» Correct Answer - D

E=dV/dx ......................................eq 1

here E=1.5 x \(10^6\) V/m

dV=10.39 V

we have to find dx

so substituing in eq 1,we get

1.5 x\(10^6\) = 10.39/dx

dx=10.39/1.5 x \(10^6\) m

so the answer is option d

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The ionisation potential of mercury is 10.39 V. How far an electron must travel in an electric field of `1.5 xx 10^(6) V//m` to gain sufficient energy to ionise mercury ?A. `(10.39)/(1.6xx10^(-19))` mB. `(10.39)/(2xx1.6xx10^(-19))` mC. `10.39 xx 1.6 xx 10^(-19)` mD. `(10.39)/(1.5xx10^(6))` m

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