1
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly

A
3.7 eV
B
3.2 eV
C
2.8 eV
D
2.5 eV
2
JEE Advanced 2013 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 $$\times$$ 108 ms$$-$$1. The final momentum of the object is

A
0.3 $$\times$$ 10$$-$$17 kg-ms$$-$$1
B
1.0 $$\times$$ 10$$-$$17 kg-ms$$-$$1
C
3.0 $$\times$$ 10$$-$$17 kg-ms$$-$$1
D
9.0 $$\times$$ 10$$-$$17 kg-ms$$-$$1
3
IIT-JEE 2009 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions $$\phi_p=2.0~\mathrm{eV}$$, $$\phi_q=2.5~\mathrm{eV}$$ and $$\phi_r=3.0~\mathrm{eV}$$, respecticely. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is (Take hc = 1240 eV nm)

A
IIT-JEE 2009 Paper 2 Offline Physics - Dual Nature of Radiation Question 3 English Option 1
B
IIT-JEE 2009 Paper 2 Offline Physics - Dual Nature of Radiation Question 3 English Option 2
C
IIT-JEE 2009 Paper 2 Offline Physics - Dual Nature of Radiation Question 3 English Option 3
D
IIT-JEE 2009 Paper 2 Offline Physics - Dual Nature of Radiation Question 3 English Option 4
4
IIT-JEE 2009 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as $$E = {{{p^2}} \over {2m}}$$. Thus, the energy of the particle can be denoted by a quantum number 'n' taking values 1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the standing wave.

Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take $$h = 6.6 \times {10^{ - 34}}$$ J-s and $$e = 1.6 \times {10^{ - 19}}$$ C.

The allowed energy for the particle for a particular value of $$n$$ is proportional to

A
$${a^{ - 2}}$$
B
$${a^{ - 3/2}}$$
C
$${a^{ - 1}}$$
D
$${a^2}$$
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