33% OFF
ExamGOAL
MOST RELEVANT

JEE Main Ultimate Online Test Series - 2027

458 Tests
10,140 Questions
English & Hindi (हिन्दी) Languages
331 Topic Tests87 Chapter Tests30 Full Tests10 Part Tests
  • Most Relevant Questions for JEE Main 2027
  • JEE Main Predictive Percentile and Rank
  • Best Solution to Every Question
  • Very Detailed Analysis
₹999 ₹1,499
Save ₹500
Access valid till 31 May 2027
Check Out
1
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

JEE Advanced Subjects

Browse all chapters by subject