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An electron is moving with an initial velocity vecv=v_(0)hati and is in a magnetic field vecB=B_(0)hatj Then it's de-Broglie wavelength |
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Answer» remains constant `vecF_(m)=(vecBxxvecv)` `=E{B_(0)hatjxxv_(0)hati}` `=eB_(0)v_(0)(hatjxxhati=eB_(0)v_(0) (-HATK))` `implies` Direction of `vecF_(m)` would be along-Z axis which is PERPENDICULAR to `VECV`.Hence,given electron would perform circular motion with constant speed `v_(0)` in a plane perpendicular to magnetic field.Hence ,its de-Broglie wavelength would be constant equal to `lambda=(h)/(mv_(0))` .In above eqution h,m ,`v_(0)` are same hence wavelength `lambda` will remian constant. |
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