When light of wavelength 248nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is …….. $A^{\circ} .$ (Round off to the Nearest Integer). [Use: $\sqrt{3} = 1.73, h = 6.63 \times 10^{- 34}$ ] Js
[ $m_{e} = 9.1 \times 10^{- 31} k g; c = 3.0 \times 10^{8} m s^{- 1}; 1 e V = 1.6 \times 10^{- 19} J$ ]

Question

When light of wavelength 248nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is …….. $A^{\circ} .$ (Round off to the Nearest Integer). [Use: $\sqrt{3} = 1.73, h = 6.63 \times 10^{- 34}$ ] Js
[ $m_{e} = 9.1 \times 10^{- 31} k g; c = 3.0 \times 10^{8} m s^{- 1}; 1 e V = 1.6 \times 10^{- 19} J$ ]

Accepted Answer

Energy of incident radiation =hcλ6.63×10−34×3×108248×10−90.0802×10−178.02×10-19J(w)work function = 3ev = 3×1.602×10−19J                                             =4.806×10-19J    K.E of photo  e- =       hv-w                                  =8.02×10-19- 4.806×10−19                                              =3.214×10−19J            debrogli  wavelength λ= h2mKE λ=6.63× 10−342×9.1×10−31×3.214×10−19        λ=6.63× 10−3458.49×10−50=6.637.64×10−9          =0.86×10−9m  λ=8.6A0

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App

When light of wavelength 248nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is …….. A∘. (Round off to the Nearest Integer). [Use: 3=1.73,h=6.63×10−34 ] Js[ me=9.1×10−31kg;c=3.0×108ms−1;1eV=1.6×10−19J ]