An $X$-ray tube operates at $50 \ kV$. It converts $1\%$ of the energy into $X$-rays. If the amount of heat produced is $495 \ W$, then the number of electrons striking the target per second is:

$\Delta \lambda$ is the difference between the wavelength of the $K_\alpha$ line and the minimum wavelength of the continuous $X$-ray spectrum when the $X$-ray tube is operated at a voltage $V$. If the operating voltage is changed to $V / 3$,then the above difference is $\Delta \lambda^{\prime}$. Then:

If the frequency of $K_\alpha$ $X$-rays emitted from the element with atomic number $31$ is $v$,then the frequency of $K_\alpha$ $X$-rays emitted from the element with atomic number $51$ would be:

In an $X$-ray tube,when the accelerating voltage is increased from $10 \ kV$ to $20 \ kV$,the difference between the wavelength of the $K_\alpha$ line and the short-wavelength limit of the continuous spectrum increases by a factor of $3$. The atomic number of the target element is:

When an $X$-ray tube operates at $60 \ kV$,the observed tube current is $50 \ mA$. Assuming that the total energy of the electrons is converted into heat,the rate of heat production at the anode in $cal/s$ is:

On increasing the number of electrons striking the anode of an $X-$ray tube,which one of the following parameters of the resulting $X-$rays would increase?