+91-8076753736[email protected]
logo
logo
-->

Wave Optics

Question
CBSEENPH12039946
Wired Faculty App

Light of wavelength 500 nm is incident on metal with work function 2.28 eV. The de - Broglie wavelength of the emitted electrons is 

  • <2.8 x 10-10 m 

  • <2.8 x 10-9 m

  • > equal  to 2.8 x 10-9 m

  • < equal to 2.8 x 10-12 m 

Solution

B.

<2.8 x 10-9 m

As, energy of photon, E = hv
straight E space space equals space hc over straight lambda rightwards double arrow space straight E space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent space straight x space 3 space straight x space 10 to the power of 8 over denominator 500 space straight x space 10 to the power of negative 9 end exponent end fraction rightwards double arrow space straight E space equals space fraction numerator 0.0397 space straight x space 10 to the power of negative 34 end exponent space straight x space 10 to the power of 8 over denominator 10 to the power of negative 9 end exponent end fraction space equals space 0.0397 space straight x space 10 to the power of negative 21 end exponent space straight J equals space fraction numerator 0.0397 space straight x space 10 to the power of negative 21 end exponent over denominator 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction space equals space 0.0248 space straight x space 10 squared space eV space equals space 2.48 space eV According space to space Einstein apostrophe straight s space photoelectric space emission comma space we space have KE subscript max space equals space straight E space minus space straight W space equals space 2.48 space minus space 2.28 space equals space 0.2 space eV For space de space minus space broglie space wavelength space of space the space emitted space electron comma straight lambda subscript straight e space min end subscript space equals space fraction numerator 12.27 over denominator square root of KE subscript max space eV end root end fraction space equals space fraction numerator 12.27 over denominator square root of 0.2 end root end fraction equals space 27.436 space straight A with straight o on top space space equals 27.436 space straight x space 10 to the power of negative 10 end exponent straight m Thus comma space minimum space wavelength space space of space the space emitted space electron space is straight lambda subscript min space equals space 2.7436 space straight x 10 to the power of negative 9 end exponent space straight m space straight lambda greater or equal than space straight lambda subscript min

Some More Questions From Wave Optics Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation