MHT CET 2021 23rd September Evening Shift | Dual Nature of Radiation Question 27 | Physics

According to de-Broglie hypothesis if an electron of mass '$$m$$' is accelerated by potential difference '$$V$$', the associated wavelength is '$$\lambda$$'. When a proton of mass '$$\mathrm{M}$$' is accelerated through potential difference $$9 \mathrm{~V}$$, then the wavelength associated with it is

$$\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$$

$$\frac{3}{2} \sqrt{\frac{m}{M}}$$

$$\frac{\lambda}{3} \sqrt{\frac{M}{m}}$$

$$\frac{3}{\lambda} \sqrt{\frac{\mathrm{M}}{\mathrm{m}}}$$

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

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When wavelength of incident radiation on the metal surface is reduced from '$$\lambda_1$$' to '$$\lambda_2$$', the kinetic energy of emitted photoelectrons is tripled. The work function of metal [$$\mathrm{h}=$$ Plank's constant, $$\mathrm{c}=$$ velocity of light]

$$\frac{\mathrm{hc}}{2}\left[\frac{3 \lambda_1-\lambda_2}{\lambda_1 \lambda_2}\right]$$

$$\frac{\mathrm{hc}}{2}\left[\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right]$$

hc $$\left[\frac{3 \lambda_1-\lambda_2}{\lambda_1 \lambda_2}\right]$$

hc $$\left[\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right]$$

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

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An electron of mass '$$m$$' and a photon have same energy '$$E$$'. The ratio of de-Broglie wavelength of electron to the wavelength of photon is $$(\mathrm{c}=$$ velocity of light)

$$c \sqrt{\frac{E}{m}}$$

$$\frac{1}{c} \sqrt{\frac{2 m}{g}}$$

$$\frac{1}{c} \sqrt{\frac{E}{2 m}}$$

$$c \sqrt{\frac{m}{E}}$$

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