AIEEE 2005 | Dual Nature of Radiation Question 167 | Physics

A photocell is illuminated by a small bright source placed $$1$$ $$m$$ away. When the same source of light is placed $${1 \over 2}$$ $$m$$ away, the number of electrons emitted by photo-cathode would

increases by a factor of $$4$$

decreases by a factor of $$4$$

increases by a factor of $$2$$

decreases by a factor of $$2$$

AIEEE 2004

MCQ (Single Correct Answer)

-1

The work function of a substance is $$4.0$$ $$eV.$$ The longest wavelength of light that can cause photo-electron emission from this substance is approximately.

$$310$$ $$nm$$

$$400$$ $$nm$$

$$540$$ $$nm$$

$$220$$ $$nm$$

AIEEE 2004

MCQ (Single Correct Answer)

-1

According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal $$Vs$$ the frequency, of the incident radiation gives as straight the whose slope

depends both on the intensity of the radiation and the metal used

depends on the intensity of the radiation

depends on the nature of the metal used

is the same for the all metals and independent of the intensity of the radiation

AIEEE 2003

MCQ (Single Correct Answer)

-1

Two identical photo-cathodes receive light of frequencies $${f_1}$$ and $${f_2}$$. If the velocities of the photo electrons (of mass $$m$$ ) coming out are respectively $${v_1}$$ and $${v_2},$$ then

$$v_1^2 - v_2^2 = {{2h} \over m}\left( {{f_1} - {f_2}} \right)$$

$${v_1} + {v_2} = {\left[ {{{2h} \over m}\left( {{f_1} + {f_2}} \right)} \right]^{1/2}}$$

$$v_1^2 + v_2^2 = {{2h} \over m}\left( {{f_1} + {f_2}} \right)$$

$${v_1} - {v_2} = {\left[ {{{2h} \over m}\left( {{f_1} - {f_2}} \right)} \right]^{1/2}}$$

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