Science, asked by Nishthaakukreja4713, 11 months ago

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface

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Answered by parthtoshatwad
1

this! Light of two different frequencies whose photons have energies 1 eV and 2.5eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratioof maximum speeds of emitted electrons will be

Light of two different frequencies whose photons have energies 1 eV and 2.5eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be


Option 1) 1:4 Option 2) 1:2 Option 3) 1:1 Option 4) 1:5

1

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A aadil.khan

Answered 8 months, 1 week ago

As we discussed in concept


Conservation of energy -


h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}


h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}



h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}


where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function


- wherein





\frac{}1{2}mV^{2}_{max}=h \nu- \theta


V_{max}=\sqrt{\frac{2(h \nu - \theta)}{m}}


\frac{V_{1}}{V_{2}}=\sqrt{\frac{h \nu_{1}- \theta}{h \nu_{2} - \theta}} = \sqrt{\frac{1-0.5}{2.5-0.5}}=\sqrt{\frac{1}{4}}=\frac{1}{2}



Option 1)


1:4


This option is incorrect.


Option 2)


1:2


This option is correct.


Option 3)


1:1


This option is incorrect.


Option 4)


1:5


This option is incorrect.

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