Question 1

Which figure shows the correct variation of applied potential difference (V) with photoelectric current (I) at two different intensities of light ( I _1 I _2 ) of same wavelengths:

Accepted Answer

Given lights are of same wavelength and stopping potential is independent on intensity. Hence stopping potential will remain same. Intensity I _2 I _1, hence saturation current corresponding to I _2 will be greater than that corresponding to I _1.

Question 2

The acceptor level of a p-type semiconductor is 6 eV . The maximum wavelength of light which can create a hole would be : Given hc =1242 eV nm .

Accepted Answer

Energy is, E = hc E = 1242 ~nm - eV = 1242 6 =207 ~nm

Question 3

When light is incident on a metal surface the maximum kinetic energy of emitted electrons

Accepted Answer

Max. K.E. =h v-W_0; so Max. K.E. v

Question 4

If the kinetic energy of a free electron doubles, it's de-Broglie wavelength changes by the factor

Accepted Answer

de-Broglie wavelength, = h p = h 2 m (K E) 1 K E If K.E is doubled, becomes 2

Question 5

A proton and an -particle are accelerated from rest by 2 ~V and 4 ~V potentials, respectively. The ratio of their de-Broglie wavelength is:

Accepted Answer

De-Broglie wavelength, aligned & = h mv = h 2 mK = h 2 mq V \ & _ _ p = 2 ~m _ p V _ p q _ p 2 ~m _ V _ q _ \ & V _ p =2 ~V and V _ =4 ~V (given) \ & m _ =4 ~m _ p and q _ =2 q _ p aligned _ _p = 1 2 1 4 4 2 = 1 4 _p: _ =4: 1

Question 6

Proton (p) and electron (e) will have same deBroglie wavelength when the ratio of their momentum is (assume, m_p=1849 ~m _ e )

Accepted Answer

De Broglie wavelength is = h mv _ p = _ e h m _ p v _ p = h m _ e v _ e m _ e v _ e = m _ p v _ p p _ e = p _ p p _ p p _ e = 1 1

Question 7

The energy levels of an atom is shown is figure. Which one of these transitions will result in the emission of a photon of wavelength 124.1 ~nm ? Given ( h =6.62 10^ -34 Js )

Accepted Answer

As E ( eV )= 1240 ( nm ) = 1240 124.1 10 eV Only is transition (D), the energy gap is 10 eV So, option (d) is correct

Question 8

The de-Broglie wavelength of an electron moving with a velocity c 3 (c=3 10^8 ~m / s ) is equal to the wavelength of photon. The ratio of the kinetic energies of electron and photon is

Accepted Answer

The de-Broglie wavelength, = h m v Here, _e= h m_e c 3 and _p= h m_p c Given, _e= _p So, h m_e c 3 = h m_p c m_e m_p =3 Ratio of KE = K_e K_p = 1 2 m_e v_e^2 1 2 m_p v_p^2 = 1 3 ( m_e m_p =3, v_e= c 3 . and .v_p=c )

Question 9

Monochromatic light of wavelength 667 ~nm is produced by a helium neon laser. The power emitted is 9 ~mW . The number of photons arriving per sec on the average at a target irradiated by this beam is:

Accepted Answer

=667 10^ -9 m, P =9 10^ -3 ~W P = Nhc , N = No. of photons emitted/sec. N = 9 10^ -3 667 10^ -9 6.6 10^ -34 3 10^ 8 =3 10^ 16 / sec

Question 10

Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 ~V . The de - Broglie wavelengths are _ e and _ p for electrons and protons resp

Accepted Answer

Energy = charge potential diff. E _ electron = q _ e V and E _ proton = q _ p 4 ~V de - Broglie wavelength = h P = h 2 mE _ e = h 2 m_ e eV and _ P = h 2 m_ p e 4 ~V (q_ e =q_ p ) _ e _ P = h 2 m_ e eV h 2 m_ p e 4 ~V =2 m_ P m_ e

Question 11

A plane electromagnetic wave is incident on a plane surface of area A, normally and is perfectly reflected. If energy E strikes the surface in time t then force exerted on the surface is ( c= speed of

Accepted Answer

Incident momentum, p= E c For perfectly reflecting surface with normal incidence p=2 p= 2 E c ; F = p t = 2 E c t

Dual Nature of Matter — Previous Year Questions & Solutions