Cathode rays enter into a uniform magnetic field perpendicular to the direction of the field. In the magnetic field,their path will be:

To emit an electron from a metal surface, the minimum electric field required is . . . . . . .

$(a)$ Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500\;V$ with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron,$i.e.$,the $e/m$ is given to be $1.76 \times 10^{11}\; C\; kg^{-1}$.
$(b)$ Use the same formula you employ in $(a)$ to obtain electron speed for a collector potential of $10\;MV$. Do you see what is wrong? In what way is the formula to be modified?

In a Thomson set-up for the determination of $e/m$,electrons accelerated by $2.5 \ kV$ enter the region of crossed electric and magnetic fields of strengths $3.6 \times 10^4 \ Vm^{-1}$ and $1.2 \times 10^{-3} \ T$ respectively and go through undeflected. The measured value of $e/m$ of the electron is equal to

Positive rays were discovered by

$A$ metallic surface with a work function of $2 \, eV$,on heating to a temperature of $800 \, K$,gives an emission current of $1 \, mA$. If another metallic surface having the same surface area,same emission constant,but a work function of $4 \, eV$ is heated to a temperature of $1600 \, K$,then the emission current will be......$mA$.