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

 

Answers (1)
A Aadil Khan

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.

Exams
Articles
Questions