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  4. For photoelectric emission, tungsten requires light of 2300 mathringA . If light of 1800 mathringA wavelength is incident then emission

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Q. For photoelectric emission, tungsten requires light of 2300 $\mathring{A}$ . If light of 1800 $\mathring{A}$ wavelength is incident then emission

Punjab PMETPunjab PMET 2006Dual Nature of Radiation and Matter

Solution:

If kinetic energy of photoelectrons emitted from metal surface is $ E_k $ and $W$ is the energy required to eject photoelectrons from the metal, then from Einstein's photoelectric equation
$ E_k = hv - W $
Also, $ E = \frac{ hc }{ \lambda } $
where $h$ is Planck's constant and $c$ is speed of light and ${\lambda } $ is wavelength.
Since wavelength is reduced from $2300\, \mathring{A}$ to $1800\, \mathring{A}$, energy increases hence emission takes place.