Radiation from hydrogen gas excited to first excited state is used for illuminating certain metallic plate. When the same plate is exposed to the radiation from some unknown hydrogen like gas excited to the same level it is found that de-Broglie wavelength of fastest photoelectron has decreased by 2.3 times. It is given that energy corresponding to longest wavelength of Lyman series of the unknown gas is 3 times the ionization energy of hydrogen gas (13.6eV). The work function (in eV) of the metallic plate is ______. [Take (2.3)² = 5.25]

Based on theory

 ${}_{}\frac{}{}$ $K.E_1=\frac{1230}{800}-?$ - (i)

${}_{}{}_{}\frac{}{}$  $K.E_2=2K.E_1=\frac{1230}{500}-?$ - (ii)

from (i) & (ii)

$$ $?=0.615$  ev

hu = hu₀ + K.E

Cases u = 2u₀

h2u₀ = hu₀ + K.E₁

K.E₁ = hu₀

$\frac{1}{2}m{v}_1^2=h{\upsilon }_0$

$v_1^2=\frac{2h{\upsilon }_0}{m}$  

$v_1=\sqrt[]{\frac{2h{\upsilon }_0}{m}}$ - (1)      

Now, cases 2

h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2}m{v}_2^2=4h{\upsilon }_0$

v₂ =     $\sqrt[]{\frac{8h{\upsilon }_0}{m}}$   - (2)

$\frac{v_2}{v_1}=\sqrt[]{\frac{8}{2}=2}$          

v₂ = 2v₁

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}f\left(x\right)=\frac{2x^2-3x-2}{x-2}\\ =\frac{2x^2-4x+x-2}{x-2}=\frac{2x\left(x-2\right)+1\left(x-2\right)}{x-2}\\ =\frac{\left(2x+1\right)\left(x-2\right)}{x-2}=2x+1\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=2x+1=\underset{h\to 0}{lim}2\left(2-h\right)+1=4+1=5\\ \underset{x\to 2^{+}}{lim}f\left(x\right)=2x+1=\underset{h\to 0}{lim}2\left(2+h\right)+1=4+1=5\\ \underset{x\to 2}{lim}f\left(x\right)=5\\ As\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)=5\\ Hence,f\left(x\right)iscontinuousatx=2.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- p=100W

_{$\lambda$ 1}= 1nm

_{$\lambda$ 2}= 500nm

Also n₁and n₂ represent number of photon of x-rays and visible light emitted from the two sources per sec

E/t=P=n₁hc/ $\lambda$ ₁₌n₂hc/ $\lambda$ ₂

So n₁/ $\lambda$ ₁ = n₂/ $\lambda$ ₂

So n₁/n₂= 1/500