1
IIT-JEE 2007 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

Electrons with de-Broglie wavelength $$\lambda$$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

A
$$\lambda_{0}=\frac{2 m c \lambda^{2}}{h}$$
B
$$\lambda_{0}=\frac{2 h}{m c}$$
C
$$\lambda_{0}=\frac{2 m^{2} c^{2} \lambda^{3}}{h^{2}}$$
D
$$\lambda_{0}=\lambda$$
2
IIT-JEE 2007 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the 4 $$\times$$ 4 matrix given in the ORS.

Column I Column II
(A) Transition between two atomic energy levels (P) Characteristic X-rays
(B) Electron emission from a material (Q) Photoelectric effect
(C) Mosley's law (R) Hydrogen spectrum
(D) Change of photon energy into kinetic energy of electrons (S) $$\beta$$-decay

A
A $$\to$$ (P); B $$\to$$ (Q); C $$\to$$ (P); D $$\to$$ (Q)
B
A $$\to$$ (R); B $$\to$$ (Q, S); C $$\to$$ (P); D $$\to$$ (Q, P)
C
A $$\to$$ (P, R); B $$\to$$ (Q, S); C $$\to$$ (P); D $$\to$$ (Q)
D
A $$\to$$ (P, R); B $$\to$$ (Q, S); C $$\to$$ (Q); D $$\to$$ (P)
3
IIT-JEE 2007 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

Statement 1 :

If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change.

Statement 2 :

When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy.

A
Statement 1 is True, Statement 2 is True, Statement 2 is a CORRECT explanation for Statement 1
B
Statement 1 is True, Statement 2 is True, Statement 2 is NOT a CORRECT explanation for Statement 1
C
Statement 1 is True, Statement 2 is False
D
Statement 1 is False, Statement 2 is True
4
IIT-JEE 2005 Mains
MCQ (Single Correct Answer)
+3
-1

The potential energy of a particle of mass m is given by

$$\mathrm{U}(x)=\left\{\begin{array}{cc}\mathrm{E}_{0} & 0 \leq x \leq 1 \\ 0 & x>1\end{array}\right.$$

$$\lambda_{1}$$ and $$\lambda_{2}$$ are the de Broglie wavelengths of the particle, when $$0 \leq x \leq 1$$ and $$x > 1$$, respectively. If the total energy of particle is $$2 \mathrm{E}_{0}$$, find $$\frac{\lambda_{1}}{\lambda_{2}}$$.

A
2
B
$$\sqrt2$$
C
$$\sqrt3$$
D
3

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