Question

two bulbs a nad b are emitting monochromatic light of wavelength such that a can just ionise h atoms and b can just ionise he+ ions power of a and b is 30 W and 40 W respectivelyfind ration of photos emitted​

Answer 1

Given:

Power of bulb A = 30 W

Power of bulb B = 40 W

To find:

The ratio of photons emitted per second

Calculation:

Wavelength of light required to ionise hydrogen atoms

     $\frac{1}{\lambda_{H}} =RZ^{2}[\frac{1}{n_{1}^2} -\frac{1}{n_{2}^2} ]$

     $\frac{1}{\lambda_{H}} = R(1)^{2}[\frac{1}{(1)^2} -\frac{1}{\infty^2} ]$

     $\frac{1}{\lambda_{H}} =R$      

     $\lambda_{H} =\frac{1}{R}$

Wavelength of light required to ionise helium atoms

     $\frac{1}{\lambda_{He}} =RZ^{2}[\frac{1}{n_{1}^2} -\frac{1}{n_{2}^2} ]$

     $\frac{1}{\lambda_{He}} = R(2)^{2}[\frac{1}{(1)^2} -\frac{1}{\infty^2} ]$

     $\frac{1}{\lambda_{He}} =4R$      

     $\lambda_{He} =\frac{1}{4R}$

We know the relation

     $E=\frac{nhc}{\lambda}$

Substitute the values of λ in the formula

     $E_{H}=\frac{n_{H}hc}{\frac{1}{R} }= n_{H}hcR$             ……(1)

     $E_{He}=\frac{n_{He}hc}{\frac{1}{4R} } =4n_{He}hcR$          ……(2)

Dividing equation (1) and equation (2)

     $\frac{E_{H}}{E_{He}} =\frac{n_{H}hcR}{4n_{H}hcR}$

      $\frac{30}{40} =\frac{n_{H}}{4n_{He} }$

     $\frac{n_{H}}{n_{He} } =3$

Final answer:

The ratio of photons emitted per second by bulb A to bulb B is 3