Question #50775

A photo sensitive material surface is illuminated alternatively with light of wave length 3100Å and 6200Å. It is observed that maximum speed of the photo electron in two cases are in ratio 2:1 the work function of metal is......
(hc=12400eVÅ)
1

Expert's answer

2015-02-20T09:34:17-0500

Question #50775, Physics, Nuclear Physics | for completion

A photo sensitive material surface is illuminated alternatively with light of wavelength 3100A˚3100\,\mathrm{\AA} and 6200A˚6200\,\mathrm{\AA}. It is observed that maximum speed of the photo electron in two cases are in ratio 2:1 the work function of metal is...

(hc=12400eVÅ)


λ1=3100A˚\lambda_1 = 3100\,\mathrm{\AA}λ2=6200A˚\lambda_2 = 6200\,\mathrm{\AA}hc=12400eVA˚\mathrm{h_c} = 12400\,\mathrm{eV\AA}V1V2=21\frac{V_1}{V_2} = \frac{2}{1}


A-?

Decision

Einstein's equation for the photoelectric effect: hcv=mV22+A\mathrm{h_c v} = \frac{mV^2}{2} + \mathrm{A}

V1=2V2,V_1 = 2V_2,

ν=cλ\nu = \frac{c}{\lambda}, there ν\nu - the frequency of light, c - speed of light.


{hccλ1=mV122+Ahccλ2=mV222+A\left\{ \begin{array}{l} h_c \frac{c}{\lambda_1} = \frac{mV_1^2}{2} + \mathrm{A} \\ h_c \frac{c}{\lambda_2} = \frac{mV_2^2}{2} + \mathrm{A} \end{array} \right.{hccλ1=4mV222+Ahccλ2=mV222+A\left\{ \begin{array}{l} h_c \frac{c}{\lambda_1} = \frac{4mV_2^2}{2} + \mathrm{A} \\ h_c \frac{c}{\lambda_2} = \frac{mV_2^2}{2} + \mathrm{A} \end{array} \right.{hccλ1=4mV222+Ahccλ2A=mV222\left\{ \begin{array}{l} h_c \frac{c}{\lambda_1} = \frac{4mV_2^2}{2} + \mathrm{A} \\ h_c \frac{c}{\lambda_2} - \mathrm{A} = \frac{mV_2^2}{2} \end{array} \right.hccλ1=4(hccλ2A)+Ah_c \frac{c}{\lambda_1} = 4(h_c \frac{c}{\lambda_2} - \mathrm{A}) + \mathrm{A}hccλ1=4hccλ24A+Ah_c \frac{c}{\lambda_1} = 4h_c \frac{c}{\lambda_2} - 4\mathrm{A} + \mathrm{A}3A=4hccλ2hccλ1=hcc(4λ21λ1)3\mathrm{A} = 4h_c \frac{c}{\lambda_2} - h_c \frac{c}{\lambda_1} = h_c c\left(\frac{4}{\lambda_2} - \frac{1}{\lambda_1}\right)A=hcc3(4λ21λ1)A=124001.6101931083(4620013100)=6.41011J.\begin{array}{l} \mathrm{A} = \frac{h_{c} \cdot \mathrm{c}}{3} \left(\frac{4}{\lambda_{2}} - \frac{1}{\lambda_{1}}\right) \\ \mathrm{A} = \frac{12400 \cdot 1.6 \cdot 10^{-19} \cdot 3 \cdot 10^{8}}{3} \left(\frac{4}{6200} - \frac{1}{3100}\right) = 6.4 \cdot 10^{-11} \mathrm{J}. \end{array}


The work function of metal is A=6.41011J\mathrm{A} = 6.4 \cdot 10^{-11} \mathrm{J}.


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