Question #e0081

Question

Question #e0081

1 Answer

Impact of this question

3549 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-02T15:53:16

By Einstein's equation for photoelectric effect we know

$h ν = h ν_{o} + \frac{1}{2} m v^{2}$ $hnu=hnu_o +1/2mv^2$

Where

$h \to Planck's constant = 6.626 \times 10^{- 34} J s$ $h->"Planck's constant"=6.626xx10^-34Js$

$ν \to frequency of incident radiation = 2.00 \times 10^{15} s^{- 1}$ $nu->"frequency of incident radiation"=2.00xx10^15s^-1$

$ν_{o} \to thresold frequency of incident radiation = 1.67 \times 10^{15} s^{- 1}$ $nu_o->"thresold frequency of incident radiation"=1.67xx10^15s^-1$

$m \to mass of electron = ?$ $m->"mass of electron"=?$

$v \to velocity of ejected electron = 6.84 \times 10^{5} \frac{m}{s}$ $v->"velocity of ejected electron"=6.84xx10^5m/s$

Now

$h ν = h ν_{o} + \frac{1}{2} m v^{2}$ $hnu=hnu_o +1/2mv^2$

$\Rightarrow \frac{1}{2} m v^{2} = h (ν - ν_{\circ})$ $=>1/2mv^2=h(nu-nu_@)$

$\Rightarrow m = \frac{2 h (ν - ν_{\circ})}{v^{2}}$ $=>m=(2h(nu-nu_@))/v^2$

$\Rightarrow m = \frac{2 \cdot 6.626 \cdot 10^{-34} (2 - 1.67) \cdot 10^{15}}{{6.84}^{2} \cdot 10^{10}}$ $=>m=(2*6.626*10^"-34"(2-1.67)*10^15)/(6.84^2*10^10)$

$\Rightarrow m = \frac{2 \cdot 6.626 \cdot 0.33 \cdot 10^{- 31}}{{6.84}^{2}}$ $=>m=(2*6.626*0.33*10^-31)/6.84^2$

$\Rightarrow m = 9.34 \times 10^{- 31} k g$ $=>m=9.34xx10^-31 kg$

Science

Math

Humanities

... and beyond

Question #e0081

1 Answer

Related questions

Impact of this question