1.In a photoelectric experiment, with light of wavelength λ\lambdaλ, the fastest electron has speed vvv. If the exciting wavelength is changed to 3λ/43\lambda/43λ/4, the speed of the fastest emitted electron will becomea.v3/4v\sqrt{3/4}v3/4b.v4/3v\sqrt{4/3}v4/3c.less than v4/3v\sqrt{4/3}v4/3d.greater than v4/3v\sqrt{4/3}v4/3Login to continueOnly logged in users canattempt or see the solution.