1
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength $$\lambda _A$$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons $$\lambda _B$$ = 2$$\lambda _A$$, then the work function of metal B is :
A
1.5eV
B
4eV
C
2eV
D
3eV
2
JEE Main 2020 (Online) 7th January Evening Slot
+4
-1
An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)
A
$${1 \over c}{\left( {{{2E} \over m}} \right)^{{1 \over 2}}}$$
B
$${1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$$
C
$${\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$$
D
$$c{\left( {2mE} \right)^{{1 \over 2}}}$$
3
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
The stopping potential V0 (in volt) as a function of frequency ($$\upsilon$$) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be:
(Given: Planck’s constant (h) = 6.63 × 10–34 Js, electron charges e = 1.6 × 10–19 C)
A
1.95 eV
B
2.12 eV
C
1.82 eV
D
1.66 eV
4
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is : [Given Planck's constant h = 6.6 × 10–34 Js, speed of light c = 3.0 × 108 m/s]
A
5 × 1015
B
1.5 × 1016
C
1 × 1016
D
2 × 1016
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